# 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.

$\lim_{x->1}\frac{x^{3}-1}{x - 1}$ = $\lim_{x->1}\frac{(x-1)(x^{2}+ x + 1}{(x - 1)}$

=

= $\lim_{x->1}(x^{2}+ x + 1)$

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.

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

=

$\lim_{x->-3}(x-2)$

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

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