Continuity And Differentiability

Left Hand Limit: =

= 4

Right Hand Limit: =

= 4

ƒ(2) = 4

Since, = f(2)

f is continuous at x=2.

Left Hand Limit: =

= 7

Right Hand Limit: =

= 7

ƒ(1) = 7

Since, = f(1)

f is continuous at x=1.

LHL: =

= [By middle term splitting]

=

= 5

RHL: =

= [By middle term splitting]

=

= 5

f(3) = 5

Since, = f(3)

f is continuous at x=3.

LHL: =

= [By middle term splitting]

=

= 10

RHL: =

= [By middle term splitting]

=

= 10

f(5)= 10

Since, = f(5)

f is continuous at x=5.

LHL: =

= 3

[ = n]

RHL: =

= 3

f(0)=1

Since, f(0)

f is discontinuous at x=0.

LHL: =

=

=2

= 2 x

=

RHL: =

=

=2

= 2 x

=

f(0)= 1

Since, f(0)

f is discontinuous at x=0.

LHL: =

= 4

RHL: =

= 0

f(x) is discontinuous at x=2

LHL: =

= 3

RHL: =

= 0

f(x) is discontinuous at x=0

LHL: =

= 1

RHL: =

= 1

f(x)=5x-4 [this equation is taken as equality for x=1 lies there]

f(1)= 1

Since, = f(1)

f is continuous at x=1.

LHL: =

= 1

RHL: =

= 1

f(x)=2x-3 [this equation is taken as equality for x=1 lies there]

f(2)= 1

Since, = f(2)

f is continuous at x=2.

LHL: =

= 1

RHL: =

= -1

f(x) is discontinuous at x=0

LHL: =

=

= -1

RHL: =

=

= 1

f(x) is discontinuous at x=a

LHL: =

=

=

= 0

RHL: =

=

= 0

f(0)=2

Since, f(0)

f is discontinuous at x=0.

= 0

sin is bounded function between -1 and +1.

Also, f(0)=0

Since, = f(0)

Hence, f is a continuous function.

LHL: =

=4

RHL: =

=4

f(2)=2

Since, f(2)

f is discontinuous at x=2.

LHL: = -x

=0

RHL: =

=0

f(0)=1

Since, f(0)

f is discontinuous at x=0.

Since, f(x) is continuous at x=0

= f(0)

=

x 2 =

=

Since, f(x) is continuous at x=0

= f(0)

=

=

= -4

Since, f(x) is continuous at x=2

= = f(2)

= f(2)

k = 5

Since, f(x) is continuous at x=3

=f(3)

=f(3)

=f(3)

k = 9

f is continuous at x =

= f()

= 3

= 3 [Here x = - h]

= 3

= 3

x 1 = 3

k = 6

=

As = 0 and sin) is bounded function between -1 and +1.

= 0

Also, f(0)=0

Since, = f(0)

Hence, f is a continuous function.

: LHL: =

= 2

RHL: =

= 2

f(1)= 2

Since, = f(1)

f is continuous at x=1.

: LHL: =

= 5

RHL: =

= 5

f(2)= 5

Since, = f(2)

f is continuous at x=2.

f is continuous at x=2

= = f(2)

= = 5

2a+b=5 ……. (1)

f is continuous at x=10

= = f(2)

= = 21

10a+b=21 ……. (1)

(1) - (2)

-8a = -16

a = 2

Putting a in 1

b=1

: f is continuous at x=0

=

=

=

a =

=

=

=

= x

= 1 x 2 x x 1

=

f(x)=|x-3|

Since every modulus function is continuous for all real x, f(x) is continuous at x=3.

f(x) =

To prove differentiable , we will use the following formula.

= = f(a)

L.H.L

=

=

= 1

R.H.L:

=

=

= -1

Since, L.H.L R.H.L, f(x) is not differentiable at x=5.

Given:

ƒ(x) =

Let’s calculate the limit of f(x) when x approaches 0 from the right

7(0) + 5

= 5

Therefore,

5

Let’s calculate the limit of f(x) when x approaches 0 from the left

5 – 3(0)

= 5

Therefore,

5

Also, f(0) = 5

As we can see,

f(0) = 5

Thus, we can say that f(x) is continuous function.

Given:

ƒ(x) =

Left hand limit at x = 0

sin(0) = 0

Therefore,

0

Right hand limit at x = 0

0

Therefore,

0

Also, f(0) = 0

As,

f (0) = 0

Thus, we can say that f(x) is continuous function.

Given:

ƒ(x) =

Left hand limit and x = 1

=

=

Applying L hospital’s rule

= n

Right hand limit and x = 1

=

=

Applying L hospital’s rule

= n

Also, f(x) = n at x =1

As we can see that

Thus, f(x) is continuous at x = 1

Let f(x) = sec x

Therefore, f(x) =

f(x) is not defined when cos x = 0

And cos x = 0 when, x = and odd multiples of like

Let us consider the function

f(a) = cos a and let c be any real number. Then,

= cos c - sin c

= cos c (1) – sin c (0)

Therefore,

cos c

Similarly,

f(c) = cos c

Therefore,

f(c) = cos c

So, f(a) is continuous at a = c

Similarly, cos x is also continuous everywhere

Therefore, sec x is continuous on the open interval

Let f(x) = sec |x| and a be any real number. Then,

Left hand limit at x = a

Right hand limit at x = a

Also, f(a) = sec |a|

Therefore,

f(a)

Thus, f(x) is continuous at x = a.

We know that sin x is continuous everywhere

Consider the point x = 0

Left hand limit:

= 1

Right hand limit:

= 1

Also we have,

f(0) = 2

As,

f(0)

Therefore, f(x) is discontinuous at x = 0.

Let n be any integer

[x] = Greatest integer less than or equal to x.

Some values of [x] for specific values of x

[3] = 3

[4.4] = 4

[-1.6] = -2

Therefore,

Left hand limit at x = n

= = n – 1

Right hand limit at x = n

= = n

Also, f(n) = [n] = n

As

Therefore, f(x) = [x] is discontinuous at x = n.

Given function f(x) =

Left hand limit at x = 2

= 2(2) – 1 = 3

Right hand limit at x = 2

= = 3

Also,

f(2) = = 3

As

= 3

Therefore,

The function f(x) is continuous at x = 2.

Given function is ƒ(x) =

Left hand limit at x = 0

= 0

Right hand limit at x = 0

= 0

Also,

f(0) = 1

As,

f(x) = x for other values of x expect 0 f(x) = 1,2,3,4…

Therefore,

f(x) is not continuous everywhere expect at x = 0

Given function f(x) =

Left hand limit at x = 1:

= 1 – 1 + 2 – 2

= 0

Right hand limit at x = 1:

= 1 – 1 + 2 – 2

= 0

Also, f(1) = 4

As we can see that,

Therefore,

f(x) is not continuous at x = 1

Given function f(x) = |x| + |x - 1|

A function f(x) is said to be continuous on a closed interval [a, b] if and only if,

(i) f is continuous on the open interval (a, b)

(ii)

(iii)

Let’s check continuity on the open interval (-1, 2)

As -1 < x < 2

Left hand limit:

=|-1-0| + |(-1-0) – 1|

=1 + 2

= 3

Right hand limit:

=|2| + |2 – 1|

=2+1

= 3

Left hand limit = Right hand limit

Here a = -1 and b = 2

Therefore,

= |-1 + 0| + |(-1 + 0) - 1|

= |- 1| + |-1 – 1|

= 1 + 2 = 3

Also f(-1) = |-1| + |-1 - 1| = 1 + 2 = 3

Now,

= |2 - 0| + |(2 - 0) - 1|

= | 2| + |2 – 1|

= 2 + 1 = 3

Also f(2) = |2| + |2 - 1| = 2 + 1 = 3

Therefore,

f(x) is continuous on the closed interval [-1, 2].

Given:

f(x) = x³

If a function is differentiable at a point, it is necessarily continuous at that point.

Left hand derivative (LHD) at x = 3

Right hand derivative (RHD) at x = 3

LHD = RHD

Therefore, f(x) is differentiable at x = 3.

Also, f(3) =27

Therefore, f(x) is also continuous at x = 3.

Given function f(x) = (x-1)^1/3

LHD at x = 1

= Not defined

RHD at x = 1

= Not defined

Since, LHD and RHD doesn’t exists

Therefore, f(x) is not differentiable at x = 1.

Let a be any constant number.

Then, f(x) = a

We know that coefficient of a linear function is

Since our function is constant, y₁ = y₂

Therefore, a = 0

Now,

Thus, the derivative of a constant function is always 0.

Left hand limit at x = 5

Right hand limit at x = 5

Also f(5) = |5 – 5 |= 0

As,

= f(5)

Therefore, f(x) is continuous at x = 5

Now, lets see the differentiability of f(x)

LHD at x = 5

RHD at x = 5

Since, LHD ≠ RHD

Therefore,

f(x) is not differentiable at x = 5

Left hand limit at x = 1

f(x) = x is polynomial function and a polynomial function is continuous everywhere

Right hand limit at x = 1

f(x) = 2 - x is polynomial function and a polynomial function is continuous everywhere

Also, f(1) =1

As we can see that,

Therefore,

f(x) is continuous at x =1

Now,

LHD at x = 1

RHD at x = 1

As, LHD ≠ RHD

Therefore,

f(x) is not differentiable at x = 1

Left hand limit at x = 2

Right hand limit at x = 2

As left hand limit ≠ right hand limit

Therefore, f(x) is not continuous at x = 2

Lets see the differentiability of f(x):

LHD at x = 2

RHD at x = 2

As, LHD ≠ RHD

Therefore,

f(x) is not derivable at x = 2

Given function f(x) =

Left hand limit at x = 1:

Right hand limit at x = 1:

Also, f(1) = 1² – 1 = 0

As,

Therefore,

f(x) is continuous at x = 1

Now, let’s see the differentiability of f(x):

LHD at x = 2:

RHD at x = 2:

= 2 + 2 = 4

As, LHD ≠ RHD

Therefore,

f(x) is not differentiable at x = 2

Given function f(x) =

LHD at x = 0:

RHD at x = 0:

As, LHD ≠ RHD

Therefore,

f(x) is not differentiable at x = 0

Given function is f(x) = |x|

LHD at x = 2:

RHD at x = 2:

As, LHD = RHD

Therefore, f(x) = |x| is differentiable at x = 2

Now f’(2) =

Therefore,

f’(2) = 1

It is given that f(x) is differentiable at each x є R

For x ≤ 1,

f(x) = x² + 3x + a i.e. a polynomial

for x > 1,

f(x) = bx + 2, which is also a polynomial

Since, a polynomial function is everywhere differentiable. Therefore, f(x) is differentiable for all x > 1 and for all x < 1.

f(x) is continuous at x = 1

4 + a = b + 2

a – b + 2 = 0 …(1)

As function is differentiable, therefore, LHD = RHD

LHD at x = 1:

RHD at x = 1:

As, LHD = RHD

Therefore,

5 = b

Putting b in (1), we get,

a – b + 2 = 0

a - 5 + 2 =0

a = 3

Hence,

a = 3 and b = 5