Evaluation of limits is the basic topic in Calculus. While evaluating the limit of the function nearer to the given point help us to determine the value of the function at the given point. This will help us to determine the y-co-ordinate of the graph at the given point. In evaluation of limits we have several methods, like by direct substitution, removal of holes, evaluating the limit as x tends to positive or negative infinity, evaluation of trigonometric functions for the given real value in terms of $\pi$. Evaluation of limits is very much useful in determining the continuity of the function or the graph of the function. 

While evaluating limits algebraically we come across different methods according to the function given. The following examples will help us to evaluate the limits by using five different methods.

Method 1 Direct Substitution:

Example: Evaluate $\lim_{x\to 2}$ (7 x2 - 6x + 2)

Solution : We have $\lim_{x\to 2}$ ( 7 x2 - 6x + 2 )
Substituting x= 2, we get $\lim_{x\to 2}$ ( 7( 2)2- 6(2) + 2) = 7 (4) - 12 + 2
= 28 - 10
= 18

Therefore the limit of the given function is 18.

Method 2 Factorization:

Example: Evaluate $\lim_{x\to 3}$ $\frac{(x^{2}-9)}{(x-3)}$

Solution : We have $\lim_{x\to 3}$ $\frac{(x^{2}-9)}{(x-3)}$ = $\lim_{x\to 3}$

$\frac{(x+3)(x-3)}{(x-3)}$

= $\lim_{x\to 3}$ (x + 3)

= 3 + 3 = 6

Limit of the function is 6

Method 3: Functions of the form $\lim_{x\to a}$ $\frac{(x^{n}-a^{n})}{(x-a)}$

The standard result of the above limit is n . a(n-1)

Example: Evaluate $\lim_{x\to 5}$ $\frac{(x^{4}-625)}{(x-5)}$


Solution : We have $\lim_{x\to 5}$ $\frac{(x^{4}-625)}{(x-5)}$ = $\lim_{x\to 5}$ $\frac{(x^{4}-5^{4})}{(x-5)}$

According to the above result limit of the function is n . a(n - 1) , where n = 4 and a = 5

Therefore, $\lim_{x\to 5}$ $\frac{(x^{4}-5^{4})}{(x-5)}$ = 4 . 5 ( 4 - 1 )

= 4 . 53

= 4 (125)

Limit of the function = 500

Method 4 Rationalization:

Example: Evaluate $\lim_{x\to 0}$ $\frac{\sqrt{(1+x^{2}})-\sqrt{(1-x^{2}})}{x^{2}}$

Solution: We have $\lim_{x\to 0}$ $\frac{\sqrt{(1+x^{2}})-\sqrt{(1-x^{2}})}{x^{2}}$ = $\frac{0}{0}$

Therefore, we can rationalize the numerator

We have $\lim_{x\to 0}$ $\frac{\sqrt{(1+x^{2}})-\sqrt{(1-x^{2}})}{x^{2}}$ = $\frac{\sqrt{(1+x^{2}})-\sqrt{(1-x^{2}})}{x^{2}}$ $\times$ $\frac{\sqrt{(1+x^{2}})+\sqrt{(1-x^{2}})}{\sqrt{(1+x^{2}})+\sqrt{(1-x^{2}})}$

= $\lim_{x\to 0}$ $\frac{(1+x^{2})-(1-x^{2})}{x^{2}[\sqrt{(1+x^{2}})+\sqrt{(1-x^{2}})]}$

= $\lim_{x\to 0}$ $\frac{(1+x^{2}-1+x^{2})}{x^{2}[\sqrt{(1+x^{2}})+\sqrt{(1-x^{2}})]}$

= $\lim_{x\to 0}$ $\frac{2x^{2}}{x^{2}[\sqrt{(1+x^{2}})+\sqrt{(1-x^{2}})]}$

= $\lim_{x\to 0}$ $\frac{2}{[\sqrt{(1+x^{2}})+\sqrt{(1-x^{2}})]}$

= $\frac{2}{2}$

Value of the limit = 1

Solved Examples

Question 1: Evaluate the limit $\lim_{x\to 1}$ $\frac{(x+1)}{(2x^{2}+7x+5)}$
Solution:
 
We have $\lim_{x\to 1}$ $\frac{(x+1)}{(2x^{2}+7x+5)}$ = $\frac{(1+1)}{(2(1)^{2}+7(1)+5)}$ [Substituting x = 1]

= $\frac{2}{2+7+5}$

= $\frac{2}{14}$

= $\frac{1}{7}$

 

Question 2: Evaluate $\lim_{x\to 2}$ $\frac{x^{2}(x^{2}-4)}{(x-2)}$
Solution:
 
We have $\lim_{x\to 2}$ $\frac{x^{2}(x^{2}-4)}{(x-2)}$ = $\frac{4(4-4)}{(2-2)}$ = $\frac{0}{0}$

Since we get it as 0/0 form, we shall try to factorise the numerator or denominator or both.
By factorising the numerator we get, x2 - 4 = (x + 2) (x - 2)

Therefore, $\lim_{x\to 2}$ $\frac{x^{2}(x^{2}-4)}{(x-2)}$ = $\lim_{x\to 2}$ $\frac{x^{2}\left ( x+2 \right )\left ( x-2 \right )}{\left ( x-2 \right )}$

= $\lim_{x\to 2}$ x2 (x + 2) [By eliminating the common factors (x - 2)]

= 4 (2 + 2) [Substituting x = 2]

= 4 x 4

= 16
 

Question 3: Evaluate $\lim_{x\to 2}$ $\frac{x^{n}-2^{n}}{x-2}$ = 80
Solution:
 
By the property of limits, $\lim_{x\to a}$ $\frac{x^{n}-a^{n}}{x-a}$ = n . a(n-1)

Therefore, $\lim_{x\to 2}$ $\frac{x^{n}-2^{n}}{x-2}$ = n . 2(n-1)

Therefore, we get, n . 2(n-1) = 80

=> n . 2(n-1) = 5. 24 = 5 . 25-1

=> n = 5
 

While evaluating limits at infinity, we divide the numerator and the denominator by the highest degree of the numerator and the denominator and apply the property that $\lim_{x\to \infty}$ $\frac{1}{x}$ = 0

Solved Examples

Question 1: $\lim_{x\to \infty}$ $\frac{2x^{2}+5x+3}{3x^{2}+10x+8}$
Solution:
 
Dividing the numerator and the denominator by the highest degree x2 , we get
$\lim_{x\to \infty}$ $\frac{2x^{2}+5x+3}{3x^{2}+10x+8}$ = $\lim_{x\to \infty}$ $\frac{\frac{2x^{2}+5x+3}{x^{2}}}{\frac{3x^{2}+10x+8}{x^{2}}}$

=$\lim_{x\to \infty}$ $\frac{\frac{2x^{2}}{x^{2}}+\frac{5x}{x^{2}}+\frac{3}{x^{2}}}{\frac{3x^{2}}{x^{2}}+\frac{10x}{x^{2}}+\frac{8}{x^{2}}}$

= $\lim_{x\to \infty}$ $\frac{2+\frac{5}{x}+\frac{3}{x^{2}}}{3+\frac{10}{x}+\frac{8}{x^{2}}}$

= $\lim_{x\to \infty}$ $\frac{2+0+0}{3+0+0}$

= $\frac{2}{3}$
 

Question 2: Evaluate $\lim_{x\to \infty}$ $\frac{sin\;x}{x}$
Solution:
 
We have $\lim_{x\to \infty}$ $\frac{sin\;x}{x}$

Substituting x = $\frac{1}{y}$, we have as x -> $\infty$, y -> 0.

Therefore, we get, $\lim_{x\to \infty}$ $\frac{sin\;x}{x}$ = $\lim_{y\to 0}$ $\frac{sin\; \frac{1}{y}}{\frac{1}{y}}$

= $\lim_{y\to 0}$ $y sin\frac{1}{y}$

= $\lim_{y\to 0}$ y . $\lim_{y\to 0}$ $sin\frac{1}{y}$

= 0 . $\lim_{y\to 0}$ $sin\frac{1}{y}$ [ Lim sin ($\frac{1}{y}$) as y -> 0 is not defined ]

Therefore, we have $\lim_{x\to \infty}$ $\frac{sin\;x}{x}$ = 0
 

The following rules are followed while evaluating limits.

1. $\lim_{x\to a}$ [ f (x) + g (x) ] = $\lim_{x\to a}$ f (x) + $\lim_{x\to a}$ g (x)

2. $\lim_{x\to a}$ [ f (x) $\times$ g (x) ] = $\lim_{x\to a}$ f (x) $\times$ $\lim_{x\to a}$ g (x)

3. $\lim_{x\to a}$$\frac{f(x)}{g(x)}$ = $\frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}$

Using the above rules, we follow the following procedure to evaluate the limit.
Step 1: Plug in the value of x and evaluate the value of the function at x = a. If the value of the limit is 0 or any real number or k/0, for any real number k, then these are the corresponding limit of the original function. k/0 denote that the limit is at + $\infty$ or - $\infty$ depending on the value of k is positive or negative.

Step 2: If we get 0/0 form, then factorize the numerator or the denominator or both and cancel the common factors if any. After cancelling the common factors we plug in the values of get to obtain the limits as in step 1.

Step 3: If the function is not factorisable but with radical signs we rationalise the numerator (or) the denominator depending on the irrational expression in the given function. Then we follow the procedure in step (2) and then step (1).

Step 4: If the given limit function is of the form, $\frac{x^{n}-a^{n}}{x-a}$, we use the condition that, $\lim_{x\to a}$ $\frac{x^{n}-a^{n}}{x-a}$ = n . a(n-1) .

Step 5: While evaluating the limit of the function as x tends to infinity ($\infty$), we divide the numerator and the denominator by xn where n is the highest degree of the numerator or the denominator.