Calculation of limits is a main topic in calculus, which gives us the value of the function as x approaches a particular real value. We are already familiar with different ways of calculating limits of functions, by direct substitution, eliminating holes, by substitution, rationalization etc. In this section we are going to evaluate the limit of the function as x approaches either from the left or from the right. This helps us to examine the continuity of the function at the given point. Further this also helps us if the given function is differentiable at a given point.

## Finding One Sided Limits Algebraically

Left Hand Limits: The limit of a function as approaches from left is defined as
$\lim_{x->a^{-}}$ f (x ) = $\lim_{h->0}$ f ( a - h ), where 0 < h < 1

Right Hand Limits : The limit of a function as approaches from the right is defined as
$\lim_{x->a^{+}}$ f ( x ) = $\lim_{h->0}$ f ( a + h ), where 0 < h < 1

In the above figure, we observe that the function (graph) approaches 7 as x approaches 4 from the left.
(i. e) the left hand limit of the function = 7
$\lim_{x->4^{-}}$ f (x) = 7

The function ( graph ) approaches 2 as x approaches from the right.
$\lim_{x->4^{+}}$ f (x) = 2
(i. e) the right hand limit of the function = 2
The value of the function is 4 when x = 4, meaning f (4) = 4
Continuity of a Function: A function is said to be continuous at a point x = a, if the left hand limit, right hand limits are equal to each other and each equal to f (a).
(i. e), $\lim_{x->a^{-}}$ f (x) = f (a) = $\lim_{x->a^{+}}$ f (x)

## One Sided Limits Examples

### Solved Examples

Question 1: Let f (x) = $\frac{|x|}{x}$, when x $\ne$ 0

= 0 when x = 0

a. Evaluate $\lim_{x->0^{-}}$  f ( x )

b. Evaluate $\lim_{x->0^{+}}$  f (x)
Solution:

According to the definition of absolution function,
| x | = x, if x $\ge$ 0
= -x, if x < 0
Therefore, $\lim_{x->0^{-}}$  f ( x ) = $\lim_{x->0^{-}}$  $\frac{|x|}{x}$

= $\lim_{x->0^{-}}$  $\frac{ - x }{x}$  [ since x < 0, substituting | x | = - x ]

= -1

$\lim_{x->0^{+}}$  f ( x ) = $\lim_{x->0^{+}}$  $\frac{|x|}{x}$

= $\lim_{x->0^{+}}$  $\frac{ x }{x}$  [ since x > 0, substituting | x | = x ]

= 1

Question 2: If  f (x) = 5 x - 4 , when 0 < x < 1

= 4 x3 - 3 x, when 1 < x < 2,

Evaluate,

a. $\lim_{x->1^{-}}$  f (x)

b. $\lim_{x->1^{+}}$  f (x)
Solution:

According to the definition of one sided limits,
$\lim_{x->a^{-}}$ f ( x ) = $\lim_{h->0}f(x-h)$, where 0 < h < 1
Therefore, we have $\lim_{x->1^{-}}$  f  (x ) = $\lim_{h->0}f(1-h)$

= $\lim_{h->0}$ [ 5 ( 1- h ) - 4 ]
= 5 ( 1 ) - 4
= 5 - 4
= 1

$\lim_{x->a^{+}}$  f ( x ) = $\lim_{h->0}$ f ( x + h ), where 0 < h < 1

=>  $\lim_{x->1^{+}}$  f ( x ) = $\lim_{h->0}$ f ( 1 + h )

= $\lim_{h->0}$ [ 4( 1+ h )3  - 3 ( 1 + h ) ]

= 4 - 3
= 1

Question 3: Evaluate the one sided limits of the function f (x) = sin x + cos x when x = $\pi$
Solution:

We have f ( x ) = sin x + cos x
Left Hand limit = $\lim_{x->\pi^{-}}$  f ( x )

= $\lim_{h->0}f(\pi-h)$

= $\lim_{h->0}$ sin ( $\pi$ - h ) + cos ( $\pi$ + h )

= sin $\pi$  + cos $\pi$ = 0 -1 = -1

Right Hand Limit = $\lim_{x->\pi^{+}}$  f ( x )

= $\lim_{h->0}$ f( $\pi$ + h )

= $\lim_{h->0}$ sin ( $\pi$ + h ) + cos ( $\pi$ + h )

= sin $\pi$ + cos $\pi$
= 0 -1
= -1

Question 4: If the function f ( x ) is given by, f ( x ) = 3ax + b , if x > 1

= 11 ,  if x = 1

= 5ax - 2b , if x < 1

is continuous at x = 1, find the values of a and b.
Solution:

Since the function is continuous at x = 1,  $\lim_{x->1^{-}}$ f ( x ) = f ( 1 ) = $\lim_{x->1^{+}}$ f ( x )

$\lim_{x->1^{+}}$  f ( x ) = $\lim_{x->1^{+}}$ ( 3ax + b )

= $\lim_{h->0}$ f ( 1 + h )

= $\lim_{h->0}$ [ 3a ( 1 + h ) + b ]

= $\lim_{h->0}$ [ 3a + 3ah + b ]

= 3a + b

f ( 1 ) = 11

$\lim_{x->1^{-}}$  f (x ) = $\lim_{x->1^{-}}$ ( 5ax - 2b )

= $\lim_{h->0}$ f ( 1 - h )

= $\lim_{h->0}$ [ 5a ( 1 - h ) - 2b ]

= $\lim_{h->0}$ [ 5a - 5ah - 2b ]

= 5a - 2b

Since the function is continuous, 3a + b = 11 = 5a - 2b
Therefore, we have the two equations,
3a + b = 11
and 5a - 2b = 11

Solving the two equations by the method of substitution,
3a + b = 11
=>           b = 11 - 3 a
Substituting b = 11 - 3 b in equation ( 2 ), we get,
5 a - 2 ( 11 - 3 a ) = 11
=> 5 a - 22 + 6a = 11
=>             11 a = 11 + 22 = 33

=>                 a = $\frac{33}{11}$ = 3

Substituting a = 3 in equation ( 1 ), we get,
3 ( 3 ) + b = 11
=>         9 + b = 11
=>               b = 11 - 9 = 2

Therefore, a = 3 and b = 2