# Lim (2x-1)2-9 / x+1 x->-1 How to Evaluate the limit ?

Jun 12, 2018

$- 12$

#### Explanation:

${\lim}_{x \to - 1} \frac{{\left(2 x - 1\right)}^{2} - 9}{x + 1}$

When we try to evaluate this directly we get:

${\lim}_{x \to - 1} \frac{{\left(2 x - 1\right)}^{2} - 9}{x + 1}$

$= \frac{{\left(2 \left(- 1\right) - 1\right)}^{2} - 9}{\left(- 1\right) + 1} = \frac{{\left(- 3\right)}^{2} - 9}{1 - 1} \to \frac{0}{0}$

So the limit is indeterminate. A route to evaluating is by first expanding the bracket then simplifying the fraction like so:

${\lim}_{x \to - 1} \frac{{\left(2 x - 1\right)}^{2} - 9}{x + 1}$

$= {\lim}_{x \to - 1} \frac{4 {x}^{2} - 4 x + 1 - 9}{x + 1}$

$= {\lim}_{x \to - 1} \frac{4 {x}^{2} - 4 x - 8}{x + 1}$

$= {\lim}_{x \to - 1} \frac{4 \left({x}^{2} - x - 2\right)}{x + 1}$

$= {\lim}_{x \to - 1} \frac{4 \left(x + 1\right) \left(x - 2\right)}{x + 1}$

$= {\lim}_{x \to - 1} \frac{4 \cancel{\left(x + 1\right)} \left(x - 2\right)}{\cancel{x + 1}}$

$= {\lim}_{x \to - 1} 4 \left(x - 2\right) = - 12$