limit, continuity & chapter 02 differentiability...x 0 fx o. solution: given that, . . liml h l...

L i m i t , C o n t i n u i t y & D i f f e r e n t i a b i l i t y P a g e | 1

Introduction:

In this chapter we will study about limit that is the core tool of calculus and all other calculus

concepts are based on it. A function can be undefined at a point, but we can think about what the

function "approaches" as it gets closer and closer to that point (this is the "limit"). Also the function

may be defined at a point, but it may approach a different limit. There are many, many times where the

functional value is the same as the limit at a point. Limit is used to define continuity, derivative and

integral of a function.

Limit of a function:

The number " 𝑙 " is called limit of a function 𝑓(𝑥) at a point 𝑥 = 𝑎, if 𝑥 approaches closer and

closer to " 𝑎 " from both sides and consequently 𝑓(𝑥) approaches closer and closer to " 𝑙 ". symbolically

it is written as,

𝑙𝑖𝑚𝑥→𝑎

𝑓(𝑥) = 𝑙 𝑜𝑟 𝑓(𝑥) → 𝑙 𝑎𝑠 𝑥 → 𝑎

Graphical representation of “limit of a function” at a point:

Left Hand Limit:

If the values of f x can be made as close as we like to " 𝑙 " by taking values of x sufficiently

close to "𝑎 " (but less than 𝑎) then we write,

. . lim ( )x a

L H L f x l

Right Hand Limit:

Chapter 02 Limit, Continuity &

Differentiability


If the values of f x can be made as close as we like to " 𝑙 " by taking values of x sufficiently

close to " 𝑎 " (but greater than 𝑎) then we write,

R. . lim ( )x a

H L f x l

Existence of limit of a function 𝒇(𝒙) at 𝒙 = 𝒂:

The limit of a function 𝑓(𝑥) at 𝑥 = 𝑎 that is lim ( )x a

f x

exists if

(i) . . lim ( )x a

L H L f x l

exists

(ii) R. . lim ( )x a

H L f x l

(iii) . . . .L H L R H L l

Fundamental Properties of limit:

If 𝑓(𝑥) , 𝑔(𝑥) are two functions and lim𝑥→𝑎

𝑓(𝑥) 𝑎𝑛𝑑 lim𝑥→𝑎

𝑔(𝑥) 𝑒𝑥𝑖𝑠𝑡𝑠 𝑡ℎ𝑒𝑛

(1) lim𝑥→𝑎

{ 𝑓(𝑥) ± 𝑔(𝑥) } = lim𝑥→𝑎

𝑓(𝑥) ± lim𝑥→𝑎

𝑔(𝑥)

(2) lim𝑥→𝑎

{ 𝑓(𝑥) × 𝑔(𝑥) } = lim𝑥→𝑎

𝑓(𝑥) × lim𝑥→𝑎

𝑔(𝑥)

(3) lim𝑥→𝑎

𝑓(𝑥)

𝑔(𝑥) =

lim𝑥→𝑎

𝑓(𝑥)

lim𝑥→𝑎

𝑔(𝑥) 𝑤ℎ𝑒𝑟𝑒 𝑔(𝑎) ≠ 0

(4) lim𝑥→𝑎

{ 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 × 𝑓(𝑥)} = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 × lim𝑥→𝑎

𝑓(𝑥)

(5) lim𝑥→𝑎

{ 𝑓(𝑥)}𝑛 = {lim𝑥→𝑎

𝑓(𝑥)}𝑛 𝑤ℎ𝑒𝑟𝑒 𝑛 𝜖 𝑍

(6) lim𝑥→𝑎

(𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡) = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡

Change of limit of a variable:

Left hand limit:

𝐿. 𝐻. 𝐿 = 𝑙𝑖𝑚 𝑥→𝑎−

𝑓(𝑥). Let 𝑥 + ℎ = 𝑎 and when ℎ → 0 then 𝑥 → 𝑎.

Now, 0

. . lim ( ) lim ( ),hx a

L H L f x f a h

putting value of x .

Right hand limit:

𝑅. 𝐻. 𝐿 = 𝑙𝑖𝑚 𝑥→𝑎+

𝑓(𝑥). Let 𝑥 − ℎ = 𝑎 and when ℎ → 0 then 𝑥 → 𝑎.


Now, 0

. . lim ( ) lim ( ),hx a

L H L f x f a h

putting value of x .

Problem: A function f x is defined as follows:

2

2

1

2.4 1

1 1

x when x

f x when x

x when x

, Does 1

limx

f x

exist ?

Solution: Given that,

2

2

1

2.4 1

1 1

x when x

f x when x

x when x

1

. . limx

L H L f x

1

. . limx

R H L f x

2

1limx

x

2

1lim 1x

x

21 21 1

1 2

Since . . . .L H L R H L . So 1

limx

f x

does not exist.


2 1 0

1 0

1 0

x when x

f x when x

x when x

, Find the value of 0

limx

f x

.


2 1 0

1 0

1 0

x when x

f x when x

x when x

0

. . limx

L H L f x

0

. . limx

R H L f x

0

lim 1x

x

2

0lim 1x

x

0 1 20 1

1 1


Since . . . .L H L R H L . So 0

limx

f x

exists.

The limiting value is, 0

lim 1x

f x

.

Problem: If 1

1

1 x

f xe

then find limits from the left and the right of 0x . Does the limit of

f x at 0x exist?

Solution: Given that, 1

1

1 x

f xe

0

. . limx

L H L f x

0

. . limx

R H L f x

0

lim 0h

f h

0

lim 0h

h

0

limh

f h

0

limh

f h

10

1lim

1h he

10

1lim

1h he

1

1 0

1

1 0

Here, . .L H L and . .R H L both are exist but they are not same.

i.e, . . . .L H L R H L . So 0

limx

f x

does not exist.

Problem: If x

f xx

then find limits from the left and the right of 0x . Does the limit of f x

at 0x exist ?

Solution: Given that, x

f xx

0

. . limx

L H L f x

0

. . limx

R H L f x

0

lim 0h

f h

0

lim 0h

f h

0

limh

f h

0

limh

f h


0

limh

h

h

0limh

h

h

0

limh

h

h

0limh

h

h

1 1


i.e, . . . .L H L R H L . So 0

limx

f x

does not exist.


2

2

1 0

0 2

x

e when xf x

x when x

, Discuss the existence of 0

limx

f x

.


2

1 0

0 2

x

e when xf x

x when x

0

. . limx

L H L f x

0

. . limx

R H L f x

2

0lim

x

xe

2

0limx

x

0e 20

1 0


i.e, . . . .L H L R H L . So 0

limx

f x

does not exist.

Problem: A real function is defined by 1

xf x

x

. Find a)

1limx

f x

; b) limx

f x

and c) limx

f x


xf x

x

Solution of (a):

1

. . limx

L H L f x

1

. . limx

R H L f x


0

lim 1h

f h

0

lim 1h

f h

0

1lim

1 1h

h

h

0

1lim

1 1h

h

h

0

1limh

h

h

0

1limh

h

h

.

Here, . .L H L and . .R H L both are not exist. So 1

limx

f x

does not exist.

Solution of (b):

Let 1

xy

then 0x y

Now, 0

1lim limx y

f x fy

0

1

lim1

1y

y

y

0

1lim

1y y

1

0 1

1 .

Solution of (c):

Let 1

xy

then 0x y

Now, 0

1lim lim

x yf x f

y

0

1

lim1

1y

y

y

0

1lim

1y y

1

0 1

1 .

Exercise:



1 0

10

2

1 0

x when x

f x when x

x when x


limx

f x

.


11 2 02

11 2 02

12 12

x when x

f x x when x

x when x


2

limx

f x

.

Problem: If 1

1

1 x

f xe

then find limits from the left and the right of 0x . Does the limit of

f x at 0x exist ?

Problem: If

12

1

3x

f x

e

then find limits from the left and the right of 2x . Does the limit

of f x at 2x exist ?

Some important limits:

(1) Proof that, 0

sinlim 1x

x

x (2)

1lim 1

x

xe

x

Proof: Proof:

0

3 5

0

2 4

0

0

sin. . lim

3! 5!lim

lim 13! 5!

1 . . .

sinlim 1

x

x

x

x

xL H S

x

x xx

x

x x

R H S

x

x

2 3

3 2

3

2

1. . . lim 1

1 1 21 1 1lim 1 . . .

2! 3!

3 21 1 1lim 1 1 1 .

2! 3!

1 1 1 3 2lim 1 1 1 1

2! 3!

1 11 1

2! 3!

. .

l

x

x

x

x

x

L H Sx

x x x x xx

x x x

x x x

x x

x x x

e R H S

1

im 1

x

xe

x


(3) 1

0lim 1 x

xx e

4.

0

1lim log 1 1x

xx

Proof: Proof:

(5) 0

1lim 1

x

x

e

x

Proof:

1

0

2 3

0

3 23

0

2

0

. . lim 1

1 1 1 1 11 1 2

1lim 1 . . .

2! 3!

1 3 2

1lim 1 1 1 .

2! 3!

1 1lim 1 1 1 1 3 2

2! 3!

1 11 1

2! 3!

x

x

x

x

x

L H S x

x x x x xx x x

x

x x xx x

x x x

1

0

. .

lim 1 x

x

e R H S

x e

0

1

0

1

0

0

1. . lim log 1

lim log 1

log lim 1

log

1 . .

1lim log 1 1

x

x

x

x

x

x

L H S xx

x

x

e

R H S

xx

0

2 3

0

2 3

0

2

0

0

1. . lim

1 12! 3!

lim

2! 3!lim

lim 12! 3!

1 . .

1lim 1

x

x

x

x

x

x

x

eL H S

x

x xx

x

x xx

x

x x

R H S

e

x


Continuity:

A function f x is said to be continuous at a point x c provided the following three

conditions are satisfied:

(1) limx c

f x

exists,

(2) f c is defined,

(3) limx c

f x f c

.

Fig. (a) Continuous function. Fig. (b) Discontinuous function.

If one or more of the conditions of this definition fails to hold, then the function f x is discontinuous

at x c .


2 1 0

0 1

1 1

x when x

f x x when x

when xx

, Discus the continuity at 1x .


2 1 0

0 1

1 1

x when x

f x x when x

when xx

1

. . limx

L H L f x

1

. . limx

R H L f x

1

limx

x

1

1limx x

1 1


Here, . . . .L H L R H L . So 1

limx

f x

exists and the limiting value is, 1

lim 1x

f x

.

Now, the functional value at 1x is, 1 1f

Since, 1

lim 1x

f x f

, the given function is continuous at 1x .

Problem: If 2

1 1

3 1

x when xf x

ax when x

for what value of a, f x is continuous at 1.x


1 1

3 1

x when xf x

ax when x

1

. . limx

L H L f x

1

. . limx

R H L f x

1

lim 1x

x

2

1lim 3x

ax

2 3 a

And, the functional value at 1x is, 1 1 1f 2

Now, the given function f x will be continuous at 1x ,

if . . . . (1)L H L R H L f

, 2 3 2or a

, 3 2or a

, 3 2or a

, 1or a (Ans.)

Problem: If 2 1sin 0

0 0

x when xxf x

when x

then test the continuity at 0.x

Solution: Given that, 2 1sin 0

0 0

x when xxf x

when x

0

. . limx

L H L f x

0

. . limx

R H L f x

2

0

1lim sinx

xx

2

0

1lim sinx

xx


2

0 0

1lim .limsinx x

xx

2

0 0

1lim .limsinx x

xx

0 1 1anyvaluebetween to 0 1 1anyvaluebetween to

0 0

Here, . . . .L H L R H L . So 0

limx

f x


lim 0x

f x

.

And, the functional value at 0x is, 0 0f

Since, 0

lim 0x

f x f



0

0 1

1 1

x when x

f x x when x

x when x

, Discus the continuity at 1x .


0

0 1

1 1

x when x

f x x when x

x when x

1

. . limx

L H L f x

1

. . limx

R H L f x

1

limx

x

1

lim 1x

x

1 0

Here, . . . .L H L R H L . So 1

limx

f x

does not exist.

Hence, the given function is discontinuous at 1x .

Problem: Test the continuity of the function 2f x x x at the point 2.x

Solution: The given function is, 2f x x x

2 2

2 0 2

2 0

x x when x

x x when x

x x when x


2 2 2

( ) 2 0 2

2 2 0

x when x

f x when x

x when x

2

. . limx

L H L f x

2

. . limx

R H L f x

2

lim 2x

2

lim 2 2x

x

2 2

Here, . . . .L H L R H L . So 2

limx

f x


lim 2x

f x

.

Now, the functional value at 2x is, 2 2 2 2f 2

Since, 2

lim 2x

f x f


Exercise:


2

1 0

1 sin 02

22 2

when x

f x x when x

x when x

, Test the continuity at 02

x and

Problem: Test the continuity of the function 1 2f x x x at the point 1.x

Problem: If 1cos 0

0 0

x when xxf x

when x

then test the continuity at 0.x

Problem: If 10

2

11 12

x when xf x

x when x

then test the continuity at 1 .2

x


Differentiability of a function:

The derivative of y f x with respect to x (for any particular value of x) is denoted by 'f x

or dy

dx and defined as,

0

limx

f x x f xdy

dx x

0limx

f x h f x

h

Provided this limit exists.

Existence of Derivative:

A function y f x is called differentiable at x a if the left hand derivative and right hand

derivative at this point that is,

0

. . limh

f a h f aL H D

h

and

0

. . limh

f a h f aR H D

h

, are both exist and equal.


2 1 0

0 1

11

x when x

f x x when x

when xx

, Discuss the differentiability at 0x and 1x .


2 1 0

0 1

11

x when x

f x x when x

when xx


1st Part: For 0x ,

Since R.H.D does not exist. So the function is not differentiable at 0x .

2nd Part: For 1x ,

Since . . . .L H D R H D does not exist. So the function is not differentiable at 1x .


2

1 0

1 sin 02

22 2

when x

f x x when x

x when x

, Discuss the differentiability at 0x and 2

x

.


0

0

2 2

0

2

0

2

0

0

0 0. . lim

0lim

1 0 1lim

1 1lim

lim

lim

0

h

h

h

h

h

h

f h fL H D

h

f h f

h

h

h

h

h

h

h

h

0

0

2

0

0

0

0 0. . lim

0lim

0 1lim

1lim

1lim 1

h

h

h

h

h

f h fR H D

h

f h f

h

h

h

h

h

h

0

0

0

0

1 1. . lim

1 1lim

lim

lim 1

1

h

h

h

h

f h fL H D

h

h

h

h

h

0

0

0

0

1 1. . lim

11

1lim

1 1lim

1

1lim

1

1

1 0

1

h

h

h

h

f h fR H D

h

h

h

h

h h

h


2

1 0

1 sin 02

22 2

when x

f x x when x

x when x

1st Part: For 0x ,

Since . . . .L H D R H D does not exist. So the function is not differentiable at 0x .

2nd Part: For 2

x ,

Since . . . .L H D R H D does not exist. So the function is not differentiable at 2

x .

0

0

0

0

0

0 0. . lim

0lim

1 1 sin 0lim

1 1lim

0lim

0

h

h

h

h

h

f h fL H D

h

f h f

h

h

h

h

0

0

0

0

0

0 0. . lim

0lim

1 sinh 1 sin 0lim

1 sinh 1lim

sinhlim

1

h

h

h

h

h

f h fR H D

h

f h f

h

h

h

h

0

2

0

0

0

2 4

0

2 4

0

3

0

2 2. . lim

1 sin 22 2 2

lim

1 cosh 2lim

cosh 1lim

1 12! 4!

lim

2! 4!lim

lim2! 4!

0

h

h

h

h

h

h

h

f h fL H D

h

h

h

h

h

h h

h

h h

h

h h

0

2 2

0

2

0

2

0

0

2 2. . lim

2 22 2 2 2

lim

2 2lim

lim

lim

0

h

h

h

h

h

f h fR H D

h

h

h

h

h

h

h

h


Exercise:


2

ln 0 1

0 1 2

1 2

x when x

f x when x

x when x

, Discuss the differentiability at 1x .

Problem: Discuss the differentiability of the function 1f x x x at the point 0x and 1.x


2

3

1

1 2

1 24

x when x

f x x when x

x when x



1 0

0 1

2 1 2

x when x

f x x when x

x when x


limit, continuity & chapter 02 differentiability...x 0 fx o. solution: given that, . . liml h l...

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