# What is the value of \lim _ { x \rightarrow - 3} \frac { x ^ { 2} - 9} { x ^ { 2} - 2x - 15} ?

Apr 13, 2017

$\frac{3}{4}$

#### Explanation:

${\lim}_{x \to - 3} \frac{{x}^{2} - 9}{{x}^{2} - 2 x - 15}$

By factoring out the numerator and the denominator,

=lim_(x to -3)(cancel((x+3))(x-3))/(cancel((x+3))(x-5)) =(-3-3)/(-3-5)=(-6)/(-8)=3/4

I hope that this was clear.

Apr 13, 2017

${\lim}_{x \to - 3} \frac{{x}^{2} - 9}{{x}^{2} - 2 x - 15} = \frac{3}{4}$

#### Explanation:

Using L'Hopital's rule, we can work out the limit of an expression.

L'Hopital's rule:

${\lim}_{x \to a} \frac{f \left(x\right)}{g \left(x\right)} = {\lim}_{x \to a} \frac{f ' \left(x\right)}{g ' \left(x\right)}$

${\lim}_{x \to - 3} \frac{{x}^{2} - 9}{{x}^{2} - 2 x - 15} = {\lim}_{x \to - 3} {\left({x}^{2} - 9\right)}^{'} / {\left({x}^{2} - 2 x - 15\right)}^{'}$

$= {\lim}_{x \to - 3} \frac{2 x}{2 x - 2}$

$= {\lim}_{x \to - 3} \frac{x}{x - 1}$

${\lim}_{x \to - 3} \frac{x}{x - 1} = \frac{- 3}{- 3 - 1} = \frac{3}{4}$