Given
f(x) =
f(x) =
To find
To limit to exist, we know …….(1)
Thus to find the limit using the concept ……(2)
……..(3)
……(4)
From above equations
(from 2)
Thus, limit does not exist.
Find k so that may exist, where .
Given f(x) =
To find
To limit to exist, we know …….(1)
thus
From (1)
2(2 + 0) + 3 = (2 - 0) + k
4 + 3 = 2 + k
5 = k
Show that does not exist.
To find
To limit to exist, we know …….(1)
Thus, to find the limit using the concept ……(2)
……..(3)
……(4)
From above equations
Thus, limit does not exist.
Let f(x) be a function defined by .
Show that does not exist.
Given f(x) =
f(x) =
f(x) =
To find
To limit to exist we know …….(1)
Thus to find the limit using the concept ……(2)
……..(3)
……(4)
From above equations
Thus, limit does not exist.
Let . Prove that does not exist.
Given f(x) =
To find whether exists?
To limit to exist we know …….(1)
Thus to limit to exist ……(2)
From above equations
Thus, the limit does not exists.
Let . Prove that does not exist.
Given f(x) =
To find whether exists?
To limit to exist we know …….(1)
Thus to limit to exist ……(2)
From above equations
Thus, the limit does not exists.
Find , where
Given f(x) =
To find
To limit to exist we know …….(1)
Thus to find the limit using the concept ……(2)
……..(3)
……(4)
From above equations
Thus from (2),(3) and (4)
If . Find and .
Given f(x) =
(i)To find
To limit to exist, we know …….(1)
Thus to find the limit using the concept ……(2)
……..(3)
……(4)
…….(5)
From above equations
thus the limit exists
Thus from (5)
(ii) To find
To limit to exist, we know …….(1)
Thus to find the limit using the concept ……(2)
……..(3)
……(4)
From above equations
Thus from (2),(3) and (4)
Find , if .
Given f(x) =
To find
To limit to exist we know …….(1)
Thus to find the limit using the concept ……(2)
……..(3)
……(4)
From above equations
thus the limit does not exists
Evaluate , where
Given f(x) =
f(x) =
f(x) =
To find
To limit to exist we know …….(1)
Thus to find the limit using the concept ……(2)
……..(3)
……(4)
From above equations
Thus limit does not exists
Let a1, a2, …….an be fixed real numbers such that f(x) = (x – a1) (x – a2) …….(x – an)
What is ? For a ≠ a1, a2, …..,an compute
Given: f(x) = (x – a1)(x – a2)…..(x – an)
Now,
Find
Given f(x) =
To find
Evaluate the following one - sided limits:
Given f(x) =
To find
Evaluate the following one - sided limits:
Given f(x) =
To find
Evaluate the following one - sided limits:
Given f(x) =
To find
Evaluate the following one - sided limits:
Given f(x) =
Factorizing f(x)
f(x) =
f(x) =
f(x) =
To find
Evaluate the following one - sided limits:
Given f(x) =
To find
Evaluate the following one - sided limits:
Some standard limit are:
Thus to find:
= ∞
Evaluate the following one - sided limits:
Some standard limit are:
Thus to find:
= - ∞
Evaluate the following one - sided limits:
Given f(x) =
Factorizing f(x)
f(x) =
f(x) =
f(x) =
f(x) =
To find
Evaluate the following one - sided limits:
Evaluate the following one - sided limits:
Some standard limit are:
Thus to find:
= 2 + ∞ = ∞
Evaluate the following one - sided limits:
(xi)
Some standard limit are:
Thus to find:
= 1 - ∞ = - ∞
Show that does not exist.
Given f(x) =
To find
To limit to exist we know …….(1)
Thus to find the limit using the concept ……(2)
……..(3)
……(4)
From above equations
Thus, limit does not exist.
Find:
We know greatest integer [x] is the integer part.
For f(x) = [x]
To find:
To limit to exist we know …….(1)
Thus to find the limit using the concept ……(2)
……..(3)
……(4)
From above equations
Thus, the limit does not exist.
Find:
We know greatest integer [x] is the integer part.
For f(x) = [x]
To find:
To limit to exist we know …….(1)
Thus to find the limit using the concept ……(2)
……..(3)
……(4)
From above equations
Thus, limit does exists.
Find:
We know greatest integer [x] is the integer part.
For f(x) = [x]
To find:
To limit to exist we know …….(1)
Thus to find the limit using the concept ……(2)
……..(3)
……(4)
From above equations
Thus limit does not exists.
Prove that for all a ∈ R. Also, prove that .
To Prove:
L.H.S (Since, [a + h] = [a])
Hence, Proved.
Also,
To prove:
L.H.S (Since, [1 – h] = 0)
Hence, Proved.
Show that .
We know greatest integer [x] is the integer part.
For f(x) = x/[x]
To show
Proof:
To limit to exist we know …….(1)
Thus to find the limit using the concept ……(2)
……..(3)
……(4)
From above equations
Find . Is it equal to .
We know greatest integer [x] is the integer part.
For f(x) = x/[x]
To show
Proof:
To limit to exist we know …….(1)
Thus to find the limit using the concept ……(2)
……..(3)
……(4)
From above equations
Find .
We know greatest integer [x] is the smallest integer nearest to that number .
For f(x) = [x]
To find:
To limit to exist we know …….(1)
Thus to find the limit using the concept ……(2)
……..(3)
……(4)
From above equations
Thus limit does exists
Evaluate (if it exists), where .
Given f(x) =
To find
To limit to exist we know …….(1)
Thus to find the limit using the concept ……(2)
……..(3)
……(4)
…..(5)
From above equations
Thus the limit does not exist
Show that does not exist.
To Prove: does not exist
Let us take the left-hand limit for the function:
L.H.L
Now, multiplying and dividing by h, we get,
L.H.L
Now, taking the right-hand limit of the function, we get,
R.H.L
Now, multiplying and dividing by h, we get,
R.H.L
Clearly, L.H.L ≠ R.H.L
Hence, limit does not exist.
Let and if , find the value of k.
Let us find the limit of the function at .
Let
Therefore,
L.H.L
Now,
Hence, k = 6.
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z = (indeterminate form)
∴ we need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
Note: While modifying be careful that you don’t introduce any zero terms in the denominator
As Z
Multiplying both numerator and denominator by √(4+x)+2 so that we can remove the indeterminate form.
∴ Z
⇒ Z
{using a2 - b2 = (a + b)(a - b)}
⇒ Z
Using basic algebra of limits-
Z =
⇒ Z = 4
Use the formula:
∴ Z = 4log 5
Or,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let
∴ we need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
This question is a direct application of limits formula of exponential and logarithmic limits.
Use the formula: and
As Z =
To get the above forms, we need to divide numerator and denominator by x.
∴ Z {using basic limit algebra}
⇒ Z {using the formulae described above}
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z
∴ we need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
As Z =
∴ Z = {using (a+b)2 = a2+b2+2ab}
Using algebra of limit, we can write that
Z =
Use the formula:
∴ Z =
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z
∴ we need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
This question is a direct application of limits formula of exponential and logarithmic limits.
To get the desired forms, we need to include mx and nx as follows:
∴ Z
⇒ Z
Using algebra of limits-
Use the formula:
∴ Z =
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z
∴ we need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
This question is a direct application of limits formula of exponential and logarithmic limits.
To get similar forms as in a formula, we move as follows-
As Z
⇒ Z
Using algebra of limits we have-
Z
Use the formula:
∴
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z
∴ we need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
As Z
∴ Z
{using (a-b)2 = a2+b2-2ab}
Z
To apply the formula we need to bring the exact form present in the formula, so-
Z
{Adding and subtracting 1 in numerator}
⇒ Z
Using algebra of limits-
Z
Use the formula:
∴ Z =
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z
∴ we need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
As Z
Using Algebra of limits-
We have-
Z
Use the formula:
∴ Z = log 4 × log 2
∵ log 4 = log 22 = 2log 2
{using properties of log}
∴ Z = 2(log 2)2
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z
∴ we need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
This question is a direct application of limits formula of exponential and logarithmic limits.
To get the desired forms, we need to include mx and nx as follows:
As Z
⇒ Z
{Adding and subtracting 1 in numerator}
⇒ Z
{using algebra of limits}
To get the form as present in the formula we multiply and divide m and n into both terms respectively:
∴ Z
Use the formula:
∴ Z = m log a – n log b
{using properties of log}
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z
∴ we need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
This question is a direct application of limits formula of exponential and logarithmic limits.
To get similar forms as in a formula, we move as follows-
As Z
⇒ Z
Using algebra of limits we have-
Z
Use the formula:
∴
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z
∴ We need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
This question is a direct application of limits formula of exponential and logarithmic limits.
To get similar forms as in a formula, we move as follows-
As x→2 ∴ x-2 →0
Let x-2 = y
∴ Z =
We can’t use the formula directly as the base of log is we need to change this to e.
Applying the formula for change of base-
We have-
∴ Z =
Use the formula:
∴ Z =
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z
∴ we need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
This question is a direct application of limits formula of exponential and logarithmic limits.
To get similar forms as in formula, we move as follows-
As Z
⇒ Z
Using algebra of limits we have-
Z
Use the formula:
∴
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z
∴ We need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
This question is a direct application of limits formula of exponential and logarithmic limits.
To get similar forms as in formula, we move as follows-
Let 1/x = y
As x→∞ ⇒ y→ 0
∴ Z can be rewritten as-
Z
Use the formula:
∴ Z =
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z
∴ we need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
This question is a direct application of limits formula of exponential and logarithmic limits and also use of sandwich theorem -
To get the desired forms, we need to include mx and nx as follows:
As Z
⇒ Z {Adding and subtracting 1 in numerator}
⇒ Z
{using algebra of limits}
To get the form as present in the formula we multiply and divide x into both terms respectively:
∴ Z
{manipulating to get the forms present in formulae}
Z
Use the formula: and
∴ Z
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z =
∴ we need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
This question is a direct application of limits formula of exponential and logarithmic limits.
To get similar forms as in a formula, we move as follows-
As Z =
⇒ Z =
Using algebra of limits we have-
Z =
Use the formula:
∴
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z
∴ We need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
It also involves a trigonometric term, so there is a possibility of application of Sandwich theorem-
As Z
⇒ Z
Use the formula: and
∴ Z = log e + 1
{∵ log e = 1}
⇒ Z = 1+1 = 2
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z
∴ We need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
It also involves a trigonometric term, so there is a possibility of application of Sandwich theorem-
As Z
To get the desired form to apply the formula we need to divide numerator and denominator by x.
⇒ Z
Using algebra of limits, we have-
Z
Use the formula: and
∴ Z =
{∵ log e = 1}
⇒ Z = 2
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z
∴ We need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
It also involves a trigonometric term, so there is a possibility of application of Sandwich theorem-
As Z
To get rid of indeterminate form we will divide numerator and denominator by sin x
∴ Z
Using Algebra of limits we have-
Z
Where, A
and B
{from sandwich theorem}
As A
Let, sin x =y
As x→0 ⇒ y→0
∴ A
Using
A = log e = 1
∴
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z
∴ We need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
It also involves a trigonometric term, so there is a possibility of application of Sandwich theorem-
As Z =
Adding and subtracting 1 in the numerator to get the desired form
⇒ Z =
⇒ Z =
{using algebra of limits}
To get the desired form to apply the formula we need to divide numerator and denominator by x.
⇒ Z
Using algebra of limits, we have-
Z
Use the formula: and
∴ Z =
{∵ log e = 1}
⇒ Z = 1/2
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z
∴ We need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
As Z
To apply the formula of logarithmic limits we need to get the form that matches with one in formula
∴ We proceed as follows-
Z
⇒ Z
⇒ Z
∵ x→a ⇒ x/a →1
⇒ x/a – 1 → 0
Let, (x/a)-1 = y
∴ y→0
Hence, Z can be rewritten as-
Z
Use the formula:
∴ Z
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z =
∴ We need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
As Z =
To apply the formula of logarithmic limits we need to get the form that matches with one in formula
∴ We proceed as follows-
Z =
⇒ Z =
⇒ Z =
To apply the formula of logarithmic limit we need denominator
∴ multiplying in numerator and denominator
Hence, Z can be rewritten as-
Z =
⇒ Z =
{Using algebra of limits}
⇒ Z =
As, x→0 ⇒
Let,
∴ Z =
Use the formula:
∴ Z =
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z =
∴ We need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
As Z =
To apply the formula of logarithmic limits we need to get the form that matches with one in formula
∴ We proceed as follows-
Z =
{using properties of log}
⇒ Z =
To apply the formula of logarithmic limit, we need the x/2 denominator
∴ multiplying 1/2 in numerator and denominator
Hence, Z can be rewritten as-
Z =
⇒ Z =
{Using algebra of limits}
As x→0 ⇒
Let,
∴ Z =
Use the formula:
∴ Z =
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z =
∴ We need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
As Z =
To apply the formula of logarithmic limits we need to get the form that matches with one in formula
∴ We proceed as follows-
Z = {using properties of log}
⇒ Z =
To apply the formula of logarithmic limit, we need x/a in the denominator
∴ multiplying 1/a in numerator and denominator
Hence, Z can be rewritten as-
Z =
⇒ Z =
{Using algebra of limits}
As x→0 ⇒
Let,
∴ Z =
Use the formula:
∴ Z =
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z =
∴ We need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
As Z =
To apply the formula of logarithmic limits we need to get the form that matches with one in formula
∴ We proceed as follows-
Z =
⇒ Z =
⇒ Z =
To apply the formula of logarithmic limit we need denominator
∴ multiplying in numerator and denominator
Hence, Z can be rewritten as-
Z =
⇒ Z =
{Using algebra of limits}
⇒ Z =
As, x→0 ⇒
Let,
∴ Z =
Use the formula:
∴ Z =
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z =
∴ we need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
This question is a direct application of limits formula of exponential and logarithmic limits.
As Z =
⇒ Z =
{Adding and subtracting 1 in numerator}
⇒ Z =
{using algebra of limits}
Use the formula:
∴ Z = log 8 – log 2 =
{using properties of log}
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z =
∴ We need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
It also involves a trigonometric term, so there is a possibility of application of Sandwich theorem-
As Z =
As, 1-cos x = 2sin2(x/2)
∴ Z =
⇒ Z =
To get the desired form to apply the formula we need to divide numerator and denominator by x2.
⇒ Z =
Using algebra of limits, we have-
Z =
Use the formula: and
∴ Z =
⇒ Z = 2 log 2
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z =
∴ We need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
As Z =
To apply the formula of logarithmic limits we need to get the form that matches with one in formula
∴ multiplying numerator and denominator by
⇒ Z =
⇒ Z =
{using (a+b)(a-b)=a2-b2}
⇒ Z =
⇒ Z =
Use the formula:
∴ Z = 1/2
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z =
∴ We need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
It also involves a trigonometric term, so there is a possibility of application of Sandwich theorem-
As Z =
To apply the formula of logarithmic limits we need to get the form that matches with one in formula
∴ dividing numerator and denominator by x3
⇒ Z =
⇒ Z =
⇒ Z =
{using algebra of limits}
Use the formula: and
∴ Z = 1/1
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z =
∴ we need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
This question is a direct application of limits formula of exponential limits.
As Z =
⇒ Z =
⇒ Z =
{using properties of exponents}
⇒ Z =
{using algebra of limits}
⇒ Z =
∴ Z =
As, x→ (π/2)
∴ cot(π/2) – cos(π/2) → 0
Let, y = cot x – cos x
∴ if x→π/2 ⇒ y→0
Hence, Z can be rewritten as-
Z =
Use the formula:
∴ Z = log a
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z =
∴ We need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
It also involves a trigonometric term, so there is a possibility of application of Sandwich theorem-
As Z =
To apply the formula we need to get the form as present in the formula. So we proceed as follows-
∵ Z =
Multiplying numerator and denominator by √(1+cos x)
⇒ Z =
Using (a+b)(a-b) = a2-b2
Z =
∵ √(1-cos2x) = sin x
⇒ Z =
{using algebra of limits}
⇒ Z =
Dividing numerator and denominator by x-
Z =
⇒ Z =
Use the formula: and
∴ Z =
{∵ log e = 1}
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z =
∴ we need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
This question is a direct application of limits formula of exponential limits.
As Z =
⇒ Z =
⇒ Z =
{using properties of exponents}
⇒ Z =
{using algebra of limits}
As, x→ 5
∴ x-5→ 0
Let, y = x-5
∴ if x→5 ⇒ y→0
Hence, Z can be rewritten as-
Z =
Use the formula:
∴ Z = e5 log e
{∵ log e = 1}
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z =
∴ we need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
This question is a direct application of limits formula of exponential limits.
As Z =
⇒ Z =
⇒ Z =
{using algebra of limits}
Use the formula:
∴ Z = e2 log e
{∵ log e = 1}
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z =
∴ we need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
This question is a direct application of limits formula of exponential limits.
As x→ π/2
∴ cos x → 0
Let, y = cos x
∴ if x→ π/2 ⇒ y→0
Hence, Z can be rewritten as-
Use the formula:
∴ Z = 1
{∵ log e = 1}
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z = =
∴ we need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
This question is a direct application of limits formula of exponential limits.
As Z =
⇒ Z =
⇒ Z =
{using algebra of limits}
Use the formula: and
∴ Z = e3 log e – 1 {∵ log e = 1}
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z = =
As we got a finite value, so no need to do any modifications.
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z =
∴ we need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
This question is a direct application of limits formula of exponential and logarithmic limits.
As Z =
⇒ Z =
{Adding and subtracting 1 in numerator}
⇒ Z =
{using algebra of limits}
To get the form as present in the formula we multiply and divide 3 and 2 into both terms respectively:
⇒ Z =
Use the formula:
∴ Z = 3log e – 2log e= 3-2 = 1
{using log e = 1}
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z = =
∴ we need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
This question is a direct application of limits formula of exponential limits.
As, x→ 0
∴ tan x → 0
Let, y = tan x
∴ if x→ 0 ⇒ y→0
Hence, Z can be rewritten as-
Use the formula:
∴ Z = log e = 1
{∵ log e = 1}
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of variable at which the limiting value is asked, if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc)
Let Z = =
∴ we need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
This question is a direct application of limits formula of exponential limits.
As Z =
⇒ Z =
⇒ Z =
{using properties of exponents}
⇒ Z =
{using algebra of limits}
⇒ Z =
∴ Z =
As, x→ 0
∴ bx-sin x → 0
Let, y = bx-sin x
∴ if x→0 ⇒ y→0
Hence, Z can be rewritten as-
Z =
Use the formula:
∴ Z = log e =1
{∵ log e = 1}
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z = =
∴ we need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
This question is a direct application of limits formula of exponential limits.
∵ Z =
To get the desired form, we proceed as follows-
Dividing numerator and denominator by tan x-
⇒ Z =
Using algebra of limits-
Z =
Use the formula - (sandwich theorem)
∴ Z =
As, x→ 0
∴ tan x → 0
Let, y = tan x
∴ if x→ 0 ⇒ y→0
Hence, Z can be rewritten as-
Use the formula:
∴ Z = log e = 1
{∵ log e = 1}
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z = =
∴ we need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
This question is a direct application of limits formula of exponential limits.
As Z =
⇒ Z =
⇒ Z =
{using properties of exponents}
⇒ Z =
{using algebra of limits}
⇒ Z =
∴ Z =
As, x→ 0
∴ x-sin x → 0
Let, y = x-sin x
∴ if x→0 ⇒ y→0
Hence, Z can be rewritten as-
Z =
Use the formula:
∴ Z = log e =1
{∵ log e = 1}
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z = =
∴ we need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
This question is a direct application of limits formula of exponential limits.
As Z =
⇒ Z =
⇒ Z =
{using algebra of limits}
Use the formula:
∴ Z = 9 log 3
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z =
∴ we need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
This question is a direct application of limits formula of exponential and logarithmic limits.
To get the desired forms, we need to include mx and nx as follows:
As Z =
⇒ Z =
{using law of exponents}
⇒ Z =
{using algebra of limits}
⇒ Z =
⇒ Z =
To get the form as present in the formula we multiply and divide by 2
∴ Z =
Use the formula:
∴ Z = 2 log a
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc)
Let Z =
∴ We need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
It also involves a trigonometric term, so there is a possibility of application of Sandwich theorem-
As Z =
As, 1-cos x = 2sin2(x/2)
∴ Z =
⇒ Z =
To get the desired form to apply the formula we need to divide numerator and denominator by x2.
⇒ Z =
Using algebra of limits, we have-
Z =
Use the formula: and
∴ Z =
⇒ Z = 2 log e = 2
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z = =
∴ we need to take steps to remove this form so that we can get a finite value.
TIP: Most of the problems of logarithmic and exponential limits are solved using the formula and
As Z =
⇒ Z = {using algebra of limits}
⇒ Z =
⇒ Z = {∵ sin(x-π/2) = -cos x}
As x→π/2
∴ x-π/2→0
Let x-π/2 = y and y→0
Z can be rewritten as-
Z =
Dividing numerator and denominator by sin y to get the form present in the formula
Z =
Using algebra of limits:
Z =
Use the formula: and
∴ Z =
Hence,
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞,0∞ .. etc.)
Let Z =
As it is not taking any indeterminate form.
∴ Z = 0
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞,1∞ .. etc.)
Let Z =
As it is taking indeterminate form.
∴ we need to take steps to remove this form so that we can get a finite value.
As, Z =
⇒ Z =
Taking log both sides-
⇒ log Z =
⇒ log Z =
{∵ log am = m log a}
Now it gives us a form that can be reduced to
Dividing numerator and denominator by tan2√x –
log Z =
using algebra of limits –
Let, tan2√x = y
As x→0+⇒ y→0+
∴ A =
Use the formula -
∴ A = 1
Now, B =
⇒ B =
Use the formula -
∴ B = 2
Hence,
log Z =
⇒ loge Z = 1/2
∴ Z = e1/2
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞,1∞ .. etc.)
Let Z =
As it is taking indeterminate form-
∴ we need to take steps to remove this form so that we can get a finite value.
As, Z =
⇒ Z =
Taking log both sides-
⇒ log Z =
⇒ log Z =
{∵ log am = m log a}
Now it gives us a form that can be reduced to
log Z = {adding and subtracting 1 to cos x to get the form}
Dividing numerator and denominator by cos x – 1 to match with form in formula
∴ log Z =
using algebra of limits –
log Z =
∴ A =
Let, cos x - 1 = y
As x→0 ⇒ y→0
∴ A =
Use the formula -
∴ A = 1
Now, B =
∵ cos x – 1 = -2sin2(x/2) and sinx = 2sin(x/2)cos(x/2)
⇒ B =
∴ B = -cot 0 = ∞
∴ B = ∞
Hence,
log Z =
⇒ loge Z = 0
∴ Z = e0 = 1
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞,1∞ .. etc.)
Let Z =
As it is taking indeterminate form-
∴ we need to take steps to remove this form so that we can get a finite value.
As, Z =
⇒ Z =
Taking log both sides-
⇒ log Z =
⇒ log Z =
{∵ log am = m log a}
Now it gives us a form that can be reduced to
log Z =
{adding and subtracting 1 to cos x to get the form}
Dividing numerator and denominator by cos x + sin x– 1 to match with form in formula
∴ log Z =
using algebra of limits –
log Z =
∴ A =
Let, cos x + sin x - 1 = y
As x→0 ⇒ y→0
∴ A =
Use the formula -
∴ A = 1
Now, B =
∵ cos x – 1 = -2sin2(x/2) and sin x = 2sin(x/2)cos(x/2)
⇒ B =
⇒ B =
⇒ B =
Use the formula -
⇒ B =
∴ B = 1
Hence,
log Z =
⇒ loge Z = 1
∴ Z = e1 = e
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞,1∞ .. etc.)
Let Z =
As it is taking indeterminate form-
∴ we need to take steps to remove this form so that we can get a finite value.
As, Z =
⇒ Z =
Taking log both sides-
⇒ log Z =
⇒ log Z =
{∵ log am = m log a}
Now it gives us a form that can be reduced to
Adding and subtracting 1 to cos x to get the form-
log Z =
Dividing numerator and denominator by cos x + a sin x– 1 to match with form in formula
∴ log Z =
using algebra of limits –
log Z =
∴ A =
Let, cos x + asin x - 1 = y
As x→0 ⇒ y→0
∴ A =
Use the formula -
∴ A = 1
Now, B =
∵ cos x – 1 = -2sin2(x/2) and sin x = 2sin(x/2)cos(x/2)
⇒ B =
⇒ B =
⇒ B =
Use the formula -
⇒ B =
∴ B = 1/a
Hence,
log Z =
⇒ loge Z = a
∴ Z = ea = ea
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞,1∞ .. etc.)
Let Z =
As it is taking indeterminate form-
∴ we need to take steps to remove this form so that we can get a finite value.
Z =
Take the log to bring the term in the product so that we can solve it more easily.
Taking log both sides-
log Z =
⇒ log Z =
{∵ log am = m log a}
⇒ log Z =
{using algebra of limits}
Still, if we put x = ∞ we get an indeterminate form,
Take the highest power of x common and try to bring x in the denominator of a term so that if we put x = ∞ term reduces to 0.
∴ log Z =
⇒ log Z =
⇒ log Z =
⇒ log Z =
∴ Loge Z =
⇒ Z = 1/2
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞,1∞ .. etc.)
Let Z =
As it is taking indeterminate form-
∴ we need to take steps to remove this form so that we can get a finite value.
Z =
Take the log to bring the power term in the product so that we can solve it more easily.
Taking log both sides-
log Z =
⇒ log Z =
{∵ log am = m log a}
using algebra of limits-
⇒ log Z =
⇒ log Z =
⇒ log Z =
As, 1-cos x = 2sin2(x/2)
∴ log Z =
Let (x-1)/2 = y
As x→1 ⇒ y→0
∴ Z can be rewritten as
Log Z =
⇒ log Z =
Use the formula -
∴ log Z =
⇒ log Z =
∴ Z =
Hence,
Evaluate the following limits:
Let
Putting the limit, we get,
This is an indeterminate form, so we need to solve this limit. Taking log on both sides we get,
Now, applying L-Hospital’s rule, we get,
Applying L-hospital rule again we get,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞,1∞ .. etc.)
Let Z =
As it is taking indeterminate form-
∴ we need to take steps to remove this form so that we can get a finite value.
Z =
Take the log to bring the power term in the product so that we can solve it more easily.
Taking log both sides-
log Z =
{∵ log am = m log a}
Now it gives us a form that can be reduced to
⇒ log Z =
Dividing numerator and denominator by to get the desired form and using algebra of limits we have-
log Z =
if we assume then as x→a ⇒ y→ 0
⇒ log Z =
Use the formula-
∴ log Z =
⇒ log Z =
Now it gives us a form that can be reduced to
Try to use it. We are basically proceeding with a hit and trial attempt.
⇒ log Z =
∵ sin (A+B) = sin A cos B + cos A sin B
⇒ log Z =
⇒ log Z=
⇒ log Z =
⇒ log Z =
Use the formula-
⇒ log Z = cot a – 0
∴ log Z = cot a
∴ Z = ecot a
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞,1∞ .. etc.)
Let Z =
As it is taking indeterminate form-
∴ we need to take steps to remove this form so that we can get a finite value.
Z =
Take the log to bring the term in the product so that we can solve it more easily.
Taking log both sides-
log Z =
⇒ log Z =
{∵ log am = m log a}
⇒ log Z =
{using algebra of limits}
Still, if we put x = ∞ we get an indeterminate form,
Take highest power of x common and try to bring x in denominator of a term so that if we put x = ∞ term reduces to 0.
∴ log Z =
⇒ log Z =
⇒ log Z =
⇒ log Z =
{∵ log (3/4) is a negative value as 3/4<1}
∴ Loge Z = -∞
⇒ Z = e-∞ = 0
Hence,
Given limit
Putting the value of limits directly, i.e., x = 1, we have
Hence the value of the given limit is 1.
Evaluate the following limits:
Given limit
Putting the value of limits directly, i.e. x = 0, we have
Hence the value of the given limit is 2.
Evaluate the following limits:
Given limit
Putting the value of limits directly, i.e. x = 0, we have
Hence the value of the given limit is 0.5
Evaluate the following limits:
Given limit
Putting the values of limits directly, i.e. x = 1, we have
Hence the value of the given limit is 3.
Evaluate the following limits:
Given limit
Putting the values of limit directly, i.e. x = a, we have
Hence the value of the given limit is
Evaluate the following limits:
Given limit
Putting the values of limits directly, i.e. x = 1, we have
Hence the value of the given limit is 0.5
Evaluate the following limits:
Given limit
Putting the value of limit directly, i.e. x = 0, we have
Hence the value of the given limit is
Evaluate the following limits:
Given the limit
Always remember the limiting value of a constant (such as 4, 13, b, etc.) is the constant itself.
So, the limiting value of constant 9 is itself, i.e., 9.
Evaluate the following limits:
Given the limit
Putting the limiting value directly, i.e. x = 2, we have
Hence the value of the given limit is 1.
Evaluate the following limits:
Given limit
Putting the value of limits directly, we have
Hence the value of the given limit is 6.
Evaluate the following limits:
Given the limit
Putting the value of limits directly, i.e. x = -1, we have
Hence the value of the given limit is
Evaluate the following limits:
Given limit
Putting the value of limit directly, i.e. x = 0, we have
Hence the value of the given limit is
Evaluate the following limits:
Given limit
Putting the value of limits directly, i.e. x = 3, we have
Hence the value of the given limit is 0.
Evaluate the following limits:
Given limit
Putting the value of limits directly, i.e. x = 0, we have
The given condition d ≠ 0 is reasonable because the denominator cannot be zero.
Hence the value of the given limit is .
Since the form is indeterminant
Method 1: factorization
= 2(-5)-1
= -11
Method 2: By L hospital rule:
Differentiating numerator and denominator separately:
= 4(-5) + 9
= -11
Evaluate the following limits:
Since the form is indeterminant
Method 1: factorization
Method 2: By L hospital rule:
Differentiating numerator and denominator separately:
Evaluate the following limits:
Since the form is indeterminant
Method 1: factorization
Since a2-b2 = (a + b)(a-b)
Thus
= 32 + 32
= 18
Method 2:
By L hospital rule:
Differentiating numerator and denominator separately:
= 54
Evaluate the following limits:
Since the form is indeterminant
Method 1: factorization
Since a3-b3 = (a-b)(a2 + b2 + ab) & a2-b2 = (a + b)(a-b)
= 3
Method 2: By L hospital rule:
Differentiating numerator and denominator separately:
= 3
Evaluate the following limits:
Since the form is indeterminant
Method 1: factorization
Since a3 + b3 = (a + b)(a2 + b2 - ab)
= 1 + 1 + 1
= 3
Method 2: By L hospital rule:
Differentiating numerator and denominator separately:
= 12(-1/2)2
= 12/4
= 3
Evaluate the following limits:
Since the form is indeterminant
Method 1: factorization
Method 2: By L hospital rule:
Differentiating numerator and denominator separately:
Evaluate the following limits:
Since the form is indeterminant
Method 1: factorization
Since a2-b2 = (a + b)(a-b)
= (2 + 2)(22 + 22)
= 32
Method 2: By L hospital rule:
Differentiating numerator and denominator separately:
= 4(2)3
= 32
Evaluate the following limits:
Since the form is indeterminant
Method 1: factorization
Method 2:
By L hospital rule:
Differentiating numerator and denominator separately:
Evaluate the following limits:
Since the form is indeterminant
Method 1: factorization
Since a3 + b3 = (a + b)(a2 + b2-ab) & a2-b2 = (a + b)(a-b)
= (-1)2 + (1)2 - (-1)
= 3
Method 2: By L hospital rule:
Differentiating numerator and denominator separately:
= 3(-1)2
= 3
Evaluate the following limits:
Since the form is indeterminant
Method 1: factorization
Since a3 + b3 = (a + b)(a2 + b2-ab) & a2-b2 = (a + b)(a-b)
=
=
= 25
Method 2: By L hospital rule:
Differentiating numerator and denominator separately:
= 25
Evaluate the following limits:
Since the form is indeterminant
Method 1: factorization
Since a3 + b3 = (a + b)(a2 + b2-ab) & a2-b2 = (a + b)(a-b)
Method 2: By L hospital rule:
Differentiating numerator and denominator separately:
Evaluate the following limits:
Since the form is
Method 1: factorization
Since a3 + b3 = (a + b)(a2 + b2-ab) & a2-b2 = (a + b)(a-b)
Method 2: By L hospital rule:
Differentiating numerator and denominator separately:
Evaluate the following limits:
Since the form is
Method 1: factorization
Since a3 + b3 = (a + b)(a2 + b2-ab) & a2-b2 = (a + b)(a-b)
= 2
Method 2: By L hospital rule:
Differentiating numerator and denominator separately:
= 2
Evaluate the following limits:
Since a2-b2 = (a + b)(a-b)
= 2
Evaluate the following limits:
Hence,
Evaluate the following limits:
Evaluate the following limits:
Evaluate the following limits:
Since the form is indeterminant
Method 1: factorization
Since a3 + b3 = (a + b)(a2 + b2-ab) & a2-b2 = (a + b)(a-b)
= 2
Method 2: By L hospital rule:
Differentiating numerator and denominator separately:
= 2
Evaluate the following limits:
Since the form is indeterminant
Method 1: factorization
Since a3 + b3 = (a + b)(a2 + b2-ab) & a2-b2 = (a + b)(a-b)
Since a3 + b3 = (a + b)(a2 + b2-ab) & a2-b2 = (a + b)(a-b)
= (√4 + 2)(4 + 4)
= (2 + 2)(4 + 4)
= 32
Method 2: By L hospital rule:
Differentiating numerator and denominator separately:
= 4(4)3/2
= 32
Evaluate the following limits:
Since the form is indeterminant
Method 1: factorization
Since a3 + b3 = (a + b)(a2 + b2-ab) & a2-b2 = (a + b)(a-b)
= 2a + 0
= 2a
Method 2: By L hospital rule:
Differentiating numerator and denominator separately:
= 2(a + 0)
= 2a
Evaluate the following limits:
Since a3 + b3 = (a + b)(a2 + b2-ab) & a2-b2 = (a + b)(a-b)
= 1
Evaluate the following limits:
Evaluate the following limits:
Since a3 + b3 = (a + b)(a2 + b2-ab) & a2-b2 = (a + b)(a-b)
Evaluate the following limits:
Since a3 + b3 = (a + b)(a2 + b2-ab) & a2-b2 = (a + b)(a-b)
= 6
Evaluate the following limits:
Since the form is indeterminant
Method 1: factorization
Since a3-b3 = (a-b)(a2 + b2 + ab) & a2-b2 = (a + b)(a-b)
=
=
=
=
Method 2: By L hospital rule:
Differentiating numerator and denominator separately:
Evaluate the following limits:
Since the form is indeterminant
Method 1: factorization
By long division method
Since a3-b3 = (a-b)(a2 + b2 + ab) & a2-b2 = (a + b)(a-b)
=
=
Method2: By L hospital rule:
Differentiating numerator and denominator separately:
Evaluate the following limits:
Since the form is indeterminant
Method 1: factorization
Since a3-b3 = (a-b)(a2 + b2 + ab) & a2-b2 = (a + b)(a-b)
Method 2: By L hospital rule:
Differentiating numerator and denominator separately:
Evaluate the following limits:
Since the form is indeterminant
Method 1: factorization
Method 2: By L hospital rule:
Differentiating numerator and denominator separately:
Evaluate the following limits:
Since the form is
Method 1: factorization
By long division method
Since a3-b3 = (a-b)(a2 + b2 + ab) & a2-b2 = (a + b)(a-b)
Method 2: By L hospital rule:
Differentiating numerator and denominator separately:
Evaluate the following limits:
Since the form is
Method 1: factorization
by dividing
Since a3-b3 = (a-b)(a2 + b2 + ab) & a2-b2 = (a + b)(a-b)
Method 2: By L hospital rule:
Differentiating numerator and denominator separately:
Evaluate the following limits:
Evaluate the following limits:
Evaluate the following limits:
= 2
Evaluate the following limits:
Since the form is indeterminant
Method 1: factorization
Since a3-b3 = (a-b)(a2 + b2 + ab) & a2-b2 = (a + b)(a-b)
= 1
Method 2: By L hospital rule:
Differentiating numerator and denominator separately:
= 1
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to 0
Substituting x as 0, we get an indeterminant form of
Rationalizing the given equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as 0
We get,
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to 0
Substituting x as 0, we get an indeterminant form of
Rationalizing the given equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as 0
We get
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to 0
Substituting x as 0, we get an indeterminant form of
Rationalizing the given equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as 0
We get,
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to 0
Substituting x as 0 we get an indeterminant form of
Rationalizing the given equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as 0
We get
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to 2
Substituting x as 2, we get an indeterminant form of
Rationalizing the given equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as 2
We get
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to 3
Substituting x as 3, we get an indeterminant form of
Rationalizing the given equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as 3
We get
Given
To find: the limit of the given equation when x tends to 0
Substituting x as 0, we find that it is in non-indeterminant form so by substituting x as 0 we will directly get the answer
We get as the answer
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to 1
Substituting x as 1, we get an indeterminant form of
Rationalizing the given equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as 1
We get
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to 1
Substituting x as 1, we get an indeterminant form of
Rationalizing the given equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as 1
We get
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to 3
Substituting x as 3, we get an indeterminant form of
Rationalizing the given equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as 3
We get
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to 1
Substituting x as 1 we get an indeterminant form of
Rationalizing the given equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as 1
We get
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to 0
Substituting x as 0, we get an indeterminant form of
Rationalizing the given equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as 0
We get
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to 2
Substituting x as 2, we get an indeterminant form of
Rationalizing the given equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as 2
We get
Evaluate the following limits:
G iven
To find: the limit of the given equation when x tends to 2
Substituting x as 2, we get an indeterminant form of
Rationalizing the given equation
Formula: (a+ b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as 2
We get
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to 7
Substituting x as 7, we get an indeterminant form of
Rationalizing the given equation
Formula: (a+b) (a-b) = a2-b2
Now we can see that the indeterminant form is removed, so substituting x as 7
We get
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to 0
Substituting x as 0, we get an indeterminant form of
Rationalizing the given equation,
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as 0
We get
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to 5
Substituting x as 5, we get an indeterminant form of
Rationalizing the given equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as 5
We get
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to 1
Substituting x as 1, we get an indeterminant form of
Rationalizing the given equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as 1
We get
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to 2
Substituting x as 2, we get an indeterminant form of
Rationalizing the given equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as 2
We get
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to 1
Substituting x as 1, we get an indeterminant form of
Rationalizing the given equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as 1
We get
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to 0
Substituting x as 0, we get an indeterminant form of
Rationalizing the given equation,
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as 0
We get
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to 0
Substituting x as 0, we get an indeterminant form of
Rationalizing the given equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as 0
We get
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to 4
Substituting x as 4, we get an indeterminant form of
Rationalizing the given equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as 4
We get
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to a
Substituting x as we get an indeterminant form of
Rationalizing the given equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as a
We get
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to 0
Substituting 0 as we get an indeterminant form of
Rationalizing the given equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as 0
We get
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to 0
Substituting 0 as we get an indeterminant form of
Rationalizing the given equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as 0
We get
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to 1
Substituting 1 as we get an indeterminant form of
Rationalizing the given equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as 1
We get
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to 1
Substituting 1 as we get an indeterminant form of
Rationalizing the given equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as 1
We get
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to 0
Substituting 0 as we get an indeterminant form of
Rationalizing the given equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as 0
We get
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to 1
Substituting 1 as we get an indeterminant form of
Rationalizing the given equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as 1
We get
Evaluate the following limits:
Given
To find: the limit of the given equation when h tends to 0
Substituting 0 as we get an indeterminant form of
Rationalizing the given equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting h as 0
We get
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to √10
Re-writing the equation as
Now rationalizing the above equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to √6
Re-writing the equation as
Now rationalizing the above equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as √6
Evaluate the following limits:
Given
To find: the limit of the given equation when x tends to
Re-writing the equation as
Now rationalizing the above equation
Formula: (a + b) (a - b) = a2 - b2
Now we can see that the indeterminant form is removed, so substituting x as
We need to find the limit for:
As limit can’t be find out simply by putting x = a because it is taking indeterminate form(0/0) form, so we need to have a different approach.
Let, Z =
Note: To solve the problems of limit similar to one in our question we use the formula mentioned below which can be derived using binomial theorem.
Formula to be used:
As Z does not match exactly with the form as described above so we need to do some manipulations–
Z =
⇒ Z =
Let x + 2 = y and a+2 = k
As x → a ; y → k
∴ Z =
Use the formula:
∴ Z =
Hence,
Evaluate the following limits:
We need to find the limit for:
As limit can’t be find out simply by putting x = a because it is taking indeterminate form(0/0) form, so we need to have a different approach.
Let, Z =
Note: To solve the problems of limit similar to one in our question we use the formula mentioned below which can be derived using binomial theorem.
Formula to be used:
As Z does not match exactly with the form as described above so we need to do some manipulations–
Z =
⇒ Z =
Let x + 2 = y and a+2 = k
As x → a ; y → k
∴ Z =
Use the formula:
∴ Z =
Hence,
Evaluate the following limits:
We need to find the limit for:
As limit can be find out simply by putting x = a because it is not taking indeterminate form(0/0) form, so we will be putting x = a
Let, Z =
⇒ Z =
This can be further simplified using a3 – 1 = (a–1)(a2 + a + 1)
⇒ Z =
⇒ Z = (1+a)4 + (1+a)2 + 1
Evaluate the following limits:
We need to find the limit for:
As limit can’t be find out simply by putting x = a because it is taking indeterminate form(0/0) form, so we need to have a different approach.
Let, Z =
Note: To solve the problems of limit similar to one in our question we use the formula mentioned below which can be derived using binomial theorem.
Formula to be used:
As Z matches exactly with the form as described above so we don’t need to do any manipulations–
Z =
Use the formula:
∴ Z =
Hence,
Evaluate the following limits:
We need to find the limit for:
As limit can’t be find out simply by putting x = a because it is taking indeterminate form(0/0) form, so we need to have a different approach.
Let, Z =
Note: To solve the problems of limit similar to one in our question we use the formula mentioned below which can be derived using binomial theorem.
Formula to be used:
As Z does not match exactly with the form as described above so we need to do some manipulations–
Z =
Dividing numerator and denominator by (x–a),we get
Z =
Using algebra of limits, we have –
Z =
Use the formula:
∴ Z =
Hence,
Evaluate the following limits:
We need to find the limit for:
As limit can’t be find out simply by putting x = (–1/2) because it is taking indeterminate form(0/0) form, so we need to have a different approach.
Let, Z =
⇒ Z =
Note: To solve the problems of limit similar to one in our question we use the formula mentioned below which can be derived using binomial theorem.
Formula to be used:
As Z matches exactly with the form as described above so we don’t need to do any manipulations–
Z =
Let y = 2x
As x → –1/2 ⇒ 2x = y → –1
∴ Z =
Use the formula:
∴ Z =
Hence,
Evaluate the following limits:
We need to find the limit for:
As limit can’t be find out simply by putting x = 27 because it is taking indeterminate form(0/0) form, so we need to have a different approach.
Let, Z =
Using algebra of limits, we have–
Z =
⇒ Z = (271/3 + 3) ×
⇒ Z = 6
Note: To solve the problems of limit similar to one in our question we use the formula mentioned below which can be derived using binomial theorem.
Formula to be used:
As Z matches exactly with the form as described above so we don’t need to do any manipulations–
Z = 6
Use the formula:
∴ Z =
Hence,
Evaluate the following limits:
We need to find the limit for:
As limit can’t be find out simply by putting x = 4 because it is taking indeterminate form(0/0) form, so we need to have a different approach.
Let, Z =
Note: To solve the problems of limit similar to one in our question we use the formula mentioned below which can be derived using binomial theorem.
Formula to be used:
As Z does not match exactly with the form as described above so we need to do some manipulations–
Z =
Dividing numerator and denominator by (x–4),we get
Z =
Using algebra of limits, we have –
Z =
Use the formula:
∴ Z =
Hence,
Evaluate the following limits:
We need to find the limit for:
As limit can’t be find out simply by putting x = 1 because it is taking indeterminate form(0/0) form, so we need to have a different approach.
Let, Z =
Note: To solve the problems of limit similar to one in our question we use the formula mentioned below which can be derived using binomial theorem.
Formula to be used:
As Z does not match exactly with the form as described above so we need to do some manipulations–
Z =
Dividing numerator and denominator by (x–1),we get
Z =
Using algebra of limits, we have –
Z =
Use the formula:
∴ Z =
Hence,
Evaluate the following limits:
We need to find the limit for:
As limit can’t be find out simply by putting x = –1 because it is taking indeterminate form(0/0) form, so we need to have a different approach.
Let, Z =
Note: To solve the problems of limit similar to one in our question we use the formula mentioned below which can be derived using binomial theorem.
Formula to be used:
As Z does matches exactly with the form as described above so we don’t need to do any manipulations–
Z =
Use the formula:
∴ Z = 3(–1)3–1 = 3
Hence,
Evaluate the following limits:
We need to find the limit for:
As limit can’t be find out simply by putting x = a because it is taking indeterminate form(0/0) form, so we need to have a different approach.
Let, Z =
Note: To solve the problems of limit similar to one in our question we use the formula mentioned below which can be derived using binomial theorem.
Formula to be used:
As Z does not match exactly with the form as described above so we need to do some manipulations–
Z =
Dividing numerator and denominator by (x–a),we get
Z =
Using algebra of limits, we have –
Z =
Use the formula:
∴ Z =
Hence,
If , find the value of n.
Given,
, we need to find value of n
So we will first find the limit and then equate it with 108 to get the value of n.
We need to find the limit for:
As limit can’t be find out simply by putting x = a because it is taking indeterminate form(0/0) form, so we need to have a different approach.
Let, Z =
Note: To solve the problems of limit similar to one in our question we use the formula mentioned below which can be derived using binomial theorem.
Formula to be used:
As Z matches exactly with the form as described above so we don’t need to do any manipulations–
Z =
Use the formula:
∴ Z = n(3)n–1
According to question Z = 108
∴ n(3)n–1 = 108
To solve such equations, factorize the number into prime factors and try to make combinations such that one satisfies with the equation.
⇒ n(3)n–1 = 4× 27 = 4× (3)4–1
Clearly on comparison we have –
n = 4
if , find all possible values of a.
Given,
, we need to find value of n
So we will first find the limit and then equate it with 9 to get the value of n.
We need to find the limit for:
As limit can’t be find out simply by putting x = a because it is taking indeterminate form(0/0) form, so we need to have a different approach.
Let, Z =
Note: To solve the problems of limit similar to one in our question we use the formula mentioned below which can be derived using binomial theorem.
Formula to be used:
As Z matches exactly with the form as described above so we don’t need to do any manipulations–
Z =
Use the formula:
∴ Z = 9(a)9–1 = 9a8
According to question Z = 9
∴ 9(a)8 = 9
⇒ a8 = 1 = 18 or (–1)8
Clearly on comparison we have –
a = 1 or –1
If , find all possible values of a.
Given,
, we need to find value of n
So we will first find the limit and then equate it with 405 to get the value of n.
We need to find the limit for:
As limit can’t be find out simply by putting x = a because it is taking indeterminate form(0/0) form, so we need to have a different approach.
Let, Z =
Note: To solve the problems of limit similar to one in our question we use the formula mentioned below which can be derived using binomial theorem.
Formula to be used:
As Z matches exactly with the form as described above so we don’t need to do any manipulations–
Z =
Use the formula:
∴ Z = 5(a)5–1 = 5a4
According to question Z = 405
∴ 5(a)4 = 405
⇒ a4 = 81 = 34 or (–3)4
Clearly on comparison we have –
a = 3 or –3
If , find all possible values of a.
Given,
, we need to find value of n
So we will first find the limit and then equate it with to get the value of n.
We need to find the limit for:
As limit can’t be find out simply by putting x = a because it is taking indeterminate form(0/0) form, so we need to have a different approach.
Let, Z =
Note: To solve the problems of limit similar to one in our question we use the formula mentioned below which can be derived using binomial theorem.
Formula to be used:
As Z matches exactly with the form as described above so we don’t need to do any manipulations–
Z =
Use the formula:
∴ Z = 9(a)9–1 = 9a8
According to question Z = = 4 + 5 = 9
∴ 9(a)8 = 9
⇒ a8 = 1 = 18 or (–1)8
Clearly on comparison we have –
a = 1 or –1
If , find all possible values of a.
Given,
⇒
Note: To solve the problems of limit similar to one in our question we use the formula mentioned below which can be derived using binomial theorem.
Formula to be used:
Using the formula we have –
3a3–1 = 4(1)4–1
⇒ 3a2 = 4
⇒ a2 = 4/3
∴ a = ± (2/√3)
Evaluate the following limits:
Given:
then,
Hence,
Evaluate the following limits:
Given:
Since, then
Hence,
Evaluate the following limits:
Given:
Hence,
Evaluate the following limits:
Given:
Rationalizing the numerator we get,
Taking x common from both numerator and denominator
Hence,
Evaluate the following limits:
Given:
On rationalizing the numerator we get,
Hence,
Evaluate the following limits:
Given:
On rationalizing the numerator we get,
Taking x common from both numerator and denominator
Hence,
Evaluate the following limits:
Given:
Hence,
Evaluate the following limits:
Given:
We know that,
By putting this Formula, we get,
Hence,
Evaluate the following limits:
Given:
Hence,
Evaluate the following limits:
Given:
Rationalizing the numerator and denominator we get,
Hence,
Evaluate the following limits:
Given:
We know that,
(n + 2)! = (n + 2) × (n + 1)!
By putting the value of (n+2)! , we get
Hence,
Evaluate the following limits:
Given:
On Rationalizing the Numerator we get,
Hence, = 1
Evaluate the following limits:
Given:
On Rationalizing the numerator we get,
Dividing the numerator and the denominator by √x, we get,
Hence,
Evaluate the following limits:
Given:
Formula Used:
Now, Putting this formula and we get,
Taking x3 as common and we get,
Since, then,
Hence,
Evaluate the following limits:
Given:
Taking LCM then, we get,
Therefore,
By putting this, we get,
Hence,
Evaluate the following limits:
Given:
Here we know that,
Since,
Hence,
Evaluate the following limits:
Formula Used:
Given:
By putting this, in the given equation, we get,
Taking x4 as common,
Hence,
Evaluate the following limits:
Now, Rationalizing the Numerator, we get,
Hence,
Evaluate the following limits:
……(1)
We can see that this is a geometric progression with the common ratio 1/3.
And, we know the sum of n terms of GP is
Let suppose, and , then
Now, putting the value of Sn in equation (1), we get
Hence,
Evaluate the following limits:
, where a is a non-zero real number.
Give:
Now, Taking x4 as common from both numerator and denominator,
Hence, a = 1
Evaluate the following limits:
and , then prove that f(-2) = f(2) = 1.
Given:
To Prove: f(-2) = f(2) = 1.
Proof: we have,
And,
Therefore, b = 1
Also,
b = 1
Thus,
On substituting the value of a and b we get,
So, f(x) = 1
Then, f(-2) = 1
Also, f(2) = 1
Hence, f(2)=f(-2)=1
Show that
To Prove:
We have L.H.S
Rationalizing the numerator, we get,
Taking x as common from both numerator and denominator,
Therefore,
Now , Take R.H.S
Now then
Therefore, R.H.S = 0
So, L.H.S ≠ R.H.S
Hence,
Evaluate the following limits:
Rationalizing the numerator, we get
Taking x as common from both numerator and denominator,
Now then
Hence, .
Rationalizing the numerator, we get
Taking x as common from both numerator and denominator,
Now then
Hence, .
Evaluate:
Formula Used:
Now putting these value, we get,
Now then,
=
Evaluate:
Here We know,
By putting these value, we get,
Hence,
Multiplying and Dividing by 3:
As, x → 0 ⇒ 3x → 0
Now, put 3x = y
Formula used:
Therefore,
Hence the value of
Evaluate the following limits:
Multiplying and Dividing by
Formula used:
Therefore,
Hence, the value of
Evaluate the following limits:
As, x → 0 ⇒ x2 → 0
Formula used:
Therefore,
=1
Hence, the value of
Evaluate the following limits:
We know,
Therefore,
Formula used:
{∵ cos 0 = 1}
Hence, the value of
Evaluate the following limits:
We know,
Sin3x = 3sinx – 4 sin3x
Therefore,
Multiplying and Dividing by 3:
As, x → 0 ⇒ 3x → 0
Now, put 3x = y
Formula used:
Therefore,
= 3 × 1
= 3
Hence, the value of
Evaluate the following limits:
Multiplying and Dividing by 8x in numerator & Multiplying and Dividing by 2x in the denominator:
We know,
Therefore,
As, x → 0 ⇒ 8x → 0 & 2x → 0
Now, put 2x = y and 8x = t
Formula used:
Therefore,
= 4
Hence, the value of
Evaluate the following limits:
Multiplying and Dividing by mx in numerator & Multiplying and Dividing by nx in the denominator:
We know,
Therefore,
As, x → 0 ⇒ mx → 0 & nx → 0
Now, put mx = y and nx = t
Formula used:
Therefore,
Hence, the value of
Evaluate the following limits:
Multiplying and Dividing by 5x in numerator & Multiplying and Dividing by 3x in the denominator:
We know,
Therefore,
As, x → 0 ⇒ 5x → 0 & 3x → 0
Now, put 5x = y and 3x = t
Formula used:
Therefore,
Hence, the value of
Evaluate the following limits:
We know,
Formula used:
Therefore,
= 1
Hence, the value of
Evaluate the following limits:
Dividing numerator and denominator by x:
We know,
Therefore,
Formula used:
Therefore,
{∵ cos 0 = 1}
Hence, the value of
Evaluate the following limits:
We know,
Therefore,
Multiplying and Dividing by in numerator &
similarly byin denominator, we get,
We know,
Therefore,
Formula used:
Therefore,
Now, put values of m, n, k and l:
Hence, the value of
Evaluate the following limits:
Multiplying and dividing by 32 :
Formula used:
Therefore,
= 9
Hence, the value of
Evaluate the following limits:
We know,
cos 2x = 1 – 2 sin2 x
Multiplying and dividing by
Formula used:
Therefore,
Hence, the value of
Evaluate the following limits:
Dividing numerator and denominator by 6x:
We know,
Therefore,
As, x → 0 ⇒ 2x → 0 & 3x → 0
Put 2x = y and 3x = k;
Formula used:
Therefore,
Hence, the value of
Evaluate the following limits:
We know,
Therefore,
Multiplying and dividing by 10:
As,
x → 0 ⇒ 2x → 0 & 5x → 0
Put 2x = y and 5x = k;
Formula used:
Therefore,
=20 × 1
= 20
Hence, the value of
Evaluate the following limits:
Multiplying and Dividing by 3θ in numerator & Multiplying and Dividing by 2θ in the denominator:
We know,
Therefore,
As, x → 0 ⇒ 3θ → 0 & 2θ → 0
Now, put 3θ = y and 2θ = t
Formula used:
Therefore,
Hence, the value of
Evaluate the following limits:
We know,
cos 2x = 1 – 2 sin2 x
Formula used:
Therefore,
Hence, the value of
Evaluate the following limits:
As, x → 0 ⇒ x2→ 0 ⇒ 4x2 → 0
Put x2 = y
Formula used:
Therefore,
= 16 × (1)2
= 16
Hence, the value of
Evaluate the following limits:
Dividing numerator and denominator by x:
We know,
Therefore,
Formula used:
Therefore,
{∵ cos 0 = 1}
= 3
Hence, the value of
Evaluate the following limits:
Dividing numerator and denominator by x:
We know,
Therefore,
Formula used:
Therefore,
Hence, the value of
Evaluate the following limits:
Dividing numerator and denominator by x:
We know,
Therefore,
Formula used:
Therefore,
{∵ cos 0 = 1}
= 8
Hence, the value of
Evaluate the following limits:
We know,
Therefore,
{∵ cos 0 = 1}
= 2 × 1
= 2
Hence, the value of
Evaluate the following limits:
We know,
Therefore,
=2 × cos(4 × 0
=2 × cos 0
{∵ cos 0 = 1}
= 2 × 1
= 2
Hence, the value of
Evaluate the following limits:
We know,
Therefore,
Multiplying and dividing by 10:
As,
X → 0 ⇒ 4x → 0
Put 4x = k;
Formula used:
Therefore,
= 8 × 1
= 8
Hence, the value of
Evaluate the following limits:
Dividing numerator and denominator by x:
We know,
Therefore,
Put 3x = y:
Formula used:
Therefore,
Hence , the value of
Evaluate the following limits:
We know,
Therefore,
Formula used:
Therefore,
= 2 cos 2 × 1
= 2 cos 2
Hence, the value of
Evaluate the following limits:
We know,
(a + b)2 = a2 + b2 + 2ab
Therefore,
Now,
We get,
Formula used:
Therefore,
Hence, the value of
Evaluate the following limits:
We know,
{∵ sin2 x = 1 – cos2 x}
{∵ a2 – b2 = (a – b) (a+b)}
{∵ cos 0 = 1}
Hence, the value of
Evaluate the following limits:
We know,
We know,
We know,
Therefore,
As, x → 0 ⇒ 3x → 0 & 4x → 0
Put 2x = y & 4x = t:
Formula used:
Therefore,
= 2
Hence, the value of
Evaluate the following limits:
We know,
cos2x = 1 – 2sin2x
⇒ 2sin2x = 1 – cos2x
Dividing numerator and denominator by x2:
We know,
Therefore,
Put 3x = y & 5x = t:
Formula used:
Therefore,
Hence, the value of
Now, 1 - cos2x = 2 sin2x
Since,
Hence,
Evaluate the following limits:
Since,
Evaluate the following limits:
Evaluate the following limits:
Rationalize the numerator, we get
Hence,
Evaluate the following limits:
Hence,
Evaluate the following limits:
Hence,
Evaluate the following limits:
Since,
= -2(1 × 2) × 2 × 1
Hence,
Evaluate the following limits:
Hence,
Evaluate the following limits:
Since,
Hence,
Evaluate the following limits:
Since, 1 - cos 2x = 2sin2x
Since,
Hence,
Evaluate the following limits:
= 2 cos 3
Hence,
Evaluate the following limits:
We know that, cos2x = 1 – 2sin2x
Therefore,
[cos2x – 1 = (cosx + 1)(cosx – 1)]
= 2(1 + 0)
= 2
Hence,
Evaluate the following limits:
Since,
Hence,
Evaluate the following limits:
Hence,
Evaluate the following limits:
Hence,
Evaluate the following limits:
Hence,
Evaluate the following limits:
Since, 1 - cos2x = 2sin2 x
Hence,
Evaluate the following limits:
Hence,
Evaluate the following limits:
Hence,
Evaluate the following limits:
Hence,
Evaluate the following limits:
Since, sin 2x = 2 sinx .cos x
Hence,
Evaluate the following limits:
Hence,
Evaluate the following limits:
Given,
Hence,
Evaluate the following limits:
Given,
Now, divide by x
Hence,
Evaluate the following limits:
Given,
Hence,
Evaluate the following limits:
Given,
Since, sin3x = 3sinx - 4sin3x
= 4 × 1
Hence,
Evaluate the following limits:
Put
= 4
Hence,
Evaluate the following limits:
Given,
Taking x as common, we get
= 1
Hence,
Evaluate the following limits:
Given,
= 0
Hence,
Evaluate the following limits:
Here,
Hence,
Evaluate the following limits:
Explanation: Here,
Hence,
Evaluate the following limits:
Given,
Explanation:
Hence,
Evaluate the following limits:
If
Given,
To Find: Value of k?
Explanation: Here,
Taking k common from L.H.S and multiply and divide by k in R.H.S, we get
K2 = 1
K = ±1
Hence, The value of k is 1, - 1.
Given:
Assumption: Let y
So,
Hence,
Evaluate the following limits:
Given,
We know, sin2x = 2sin x.cos x
By putting this value, we get
Hence
Evaluate the following limits:
Given,
Here, cos2 x = 1-sin2x
By putting this we get,
Hence,
Evaluate the following limits:
Given,
[Applying the formula 1 – cos 2x = 2sin2x]
We know that, sin x = sin(π – x)
Therefore,
Hence,
Evaluate the following limits:
Given,
Hence,
Evaluate the following limits:
Given,
We know,
Hence,
Evaluate the following limits:
We have Given, If
If
Since,
Hence,
Evaluate the following limits:
We have
If
Let
Hence,
Evaluate the following limits:
Given ,
Let t = x - a
Then, as x→a, t→0
Hence,
Evaluate the following limits:
We have
Rationalise the numerator, we get
Hence,
Evaluate the following limits:
Given,
Now, rationalize the Numerator, we get,
Hence,
Evaluate the following limits:
Given,
Now,
Hence,
Evaluate the following limits:
Given,
Where
Hence,
Evaluate the following limits:
We have Given,
Now, Rationalize the Denominator
Hence,
Evaluate the following limits:
Given,
If x → π, then π – x → 0, let π – x = y
Rationalize the Numerator
Hence,
Evaluate the following limits:
We have Given,
Now, Rationalize the Denominator
Hence,
Evaluate the following limits:
we have
Evaluate the following limits:
We have Given,
Here, x → 1, then x – 1 → 0, let x – 1 = y
Hence,
Evaluate the following limits:
, where f(x) = sin 2x
Given, f(x) = sin 2x
Since,
Now,
Since,
Hence,
Evaluate the following limits:
Given,
Now,
Hence,
Evaluate the following limits:
We have Given,
Here, x → 1, then x – 1 → 0, let x – 1 = y
Hence,
Evaluate the following limits:
We have
Since, , then
Hence,
Evaluate the following limits:
Given,
As we know, tan2x = sec2x - 1
Hence,
Evaluate the following limits:
Divide and multiply by 2, we get
Now,
Hence,
Evaluate the following limits:
We have
Now,
We know, then, we get
Hence,
Evaluate the following limits:
We have
Now,
We know, then , we get
Hence,
Evaluate the following limits:
We have
Hence,
Evaluate the following limits:
We have
Hence,
Evaluate the following limits:
We have
When,
Hence,
Evaluate the following limits:
We have
Since,
By putting these , we get
Since,
Hence,
Evaluate the following limits:
We have Given,
Hence,
Evaluate the following limits:
Rationalizing we get,
As,
Therefore, y → 0,
Now,
Hence,
Evaluate the following limits:
We have Given,
if x → 1 then, x – 1 → 0 let x – 1 = y
Hence,
Evaluate the following limits:
[cosec2x – cot2x = 1]
[Applying, a2 – b2 = (a + b)(a – b)]
Hence,
Evaluate the following limits:
We have Given ,
Now,
Here, cos(a+b) =cos a.cos b – sin a.sin b
And, sin(a+b) = sin a.cos b+cos a.sin b
Hence,
Evaluate the following limits:
We have Given,
Hence,
Evaluate the following limits:
We have Given,
Hence, the answer is 1.
Evaluate the following limits:
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z
∴ we need to take steps to remove this form so that we can get a finite value.
Tip: Similar limit problems involving trigonometric ratios are mostly solved using sandwich theorem.
So to solve this problem we need to have a sin term so that we can make use of sandwich theorem.
Note: While modifying be careful that you don’t introduce any zero terms in the denominator
As,
Multiplying numerator and denominator by 1-cos x, We have-
⇒ Z =
{As 1-cos2x = sin2x}
⇒ Z =
To apply sandwich theorem, we need to have limit such that variable tends to 0 and following forms should be there
Here x→ π so we need to do modifications before applying the theorem.
As, sin (π-x) = sin x or sin (x - π) = -sin x and tan(π – x) = -tan x
∴ we can say that-
sin2x = sin2(x-π) and tan2x = tan2(x-π)
As x → π
∴ (x – π) → 0
Let us represent x - π with y
∴ Z =
Dividing both numerator and denominator by y2
Z =
⇒ Z = {Using basic limits algebra}
As,
∴ Z =
∴
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z =
∴ we need to take steps to remove this form so that we can get a finite value.
Note: While modifying be careful that you don’t introduce any zero terms in the denominator
As
∵
⇒ Z =
{Using basic limits algebra}
∵ (1- 2sin2x) = cos 2x
As, a2 – b2 = (a+b)(a-b)
⇒ Z =
Now put the value of x, we have-
∴ Z =
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc)
Let
∴ we need to take steps to remove this form so that we can get a finite value.
Note: While modifying be careful that you don’t introduce any zero terms in the denominator
As Z =
As, a2 – b2 = (a+b)(a-b)
∴ Z =
⇒ Z =
⇒ Z =
⇒ Z =
Multiplying cosec x + 2 to both numerator and denominator-
Z =
Z =
As, cosec2x – 1 = cot2 x
∴ Z =
⇒ Z =
∴
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z =
∴ we need to take steps to remove this form so that we can get a finite value.
Note: While modifying be careful that you don’t introduce any zero terms in the denominator
As
∵ cosec2x – 1 = cot2x
As, a2 – b2 = (a+b)(a-b)
Thus,
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let
∴ we need to take steps to remove this form so that we can get a finite value.
Note: While modifying be careful that you don’t introduce any zero terms in the denominator
As Z =
Multiplying numerator and denominator by √(2+cos x) + 1,we have-
Z =
⇒ Z =
{using a2 – b2 = (a+b)(a-b)}
⇒ Z =
{using basic algebra of limits}
⇒ Z = =
As, 1+cos x = 2cos2(x/2)
∴ Z =
Tip: Similar limit problems involving trigonometric ratios along with algebraic equations are mostly solved using sandwich theorem.
So to solve this problem we need to have a sin term so that we can make use of sandwich theorem.
∵ sin(π/2 – x) = cos x
∴ Z =
As x→π ⇒ π – x → 0
Let y = π – x
Z =
To apply sandwich theorem we have to get the similar form as described below-
∴ Z =
⇒ Z =
Hence,
Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)
Let Z =
∴ Z is not taking an indeterminate form.
∴ Limiting the value of Z is not defined.
Hence,