# Limits by Factorization Method

Limits by factorization method is applicable when the value of limit gives you any of these $\frac{0}{0}$, $\frac{\infty }{\infty }$ , 0 $\times \infty$ or $\infty$ - $\infty$ form then we will use the factorization method to evaluate that limit.
Indeterminate form : It is an expression involving two functions whose limits can not be determined. In other words the limit obtained by direct substitution is of the form $\frac{0}{0}$ or $\frac{\infty }{\infty }$ or 0 $\times \infty$ or $\infty$ - $\infty$.

## Examples on Limits by Factorization

1) Find the limit of $\lim_{x->1}\frac{x^{3}-1}{x - 1}$
Solution : $\lim_{x->1}\frac{x^{3}-1}{x - 1}$ if we plug in x= 1 in the given function then we will get $\frac{0}{0}$ which is an indeterminate form. So we will solve this limit by factorization method.
$\lim_{x->1}\frac{x^{3}-1}{x - 1}$ = $\lim_{x->1}\frac{(x-1)(x^{2}+ x + 1}{(x - 1)}$

= 1}\frac{\cancel{(x-1)}(x^{2}+&space;x&space;+&space;1)}{\cancel{(x&space;-&space;1)}}" target="_blank">$\lim_{x->1}\frac{\cancel{(x-1)}(x^{2}+&space;x&space;+&space;1)}{\cancel{(x&space;-&space;1)}}$

= $\lim_{x->1}(x^{2}+ x + 1)$
Now we can plug in x = 1 directly
$\lim_{x->1}\frac{x^{3}-1}{x - 1}$ = $1^{2}$ + 1 + 1

$\lim_{x->1}\frac{x^{3}-1}{x - 1}$ = 3

2) Find the limit of $\lim_{x->-3}\frac{x^{2}+ x - 6}{x + 3}$
Solution : $\lim_{x->-3}\frac{x^{2}+ x - 6}{x + 3}$ if we plug in x= -3 in the given function then we will get $\frac{0}{0}$ which is an indeterminate form. So we will solve this limit by factorization method.
$\lim_{x->-3}\frac{x^{2}+ x - 6}{x + 3}$ = $\lim_{x->-3}\frac{(x + 3)(x - 2)}{x + 3}$

= -3}\frac{\cancel{(x+3)}(x-2)}{\cancel{(x&space;-&space;1)}}" target="_blank">$\lim_{x->-3}\frac{\cancel{(x+3)}(x-2)}{\cancel{(x&space;+&space;3)}}$

$\lim_{x->-3}(x-2)$
Now we can plug in x = -3 directly

$\lim_{x->-3}\frac{x^{2}+ x - 6}{x + 3}$ = -3 -2

$\lim_{x->-3}\frac{x^{2}+ x - 6}{x + 3}$ = -5