# How do you evaluate the limit (sqrt(x+1)-2)/(x^2-9) as x approaches -3?

## Since $\sqrt{x + 1}$ is not defined at $x = - 3$, we will take limit as $x$ approaches $+ 3$ instead of $- 3$

Aug 22, 2016

$\frac{1}{24}$.

#### Explanation:

Reqd. Limit$= {\lim}_{x \rightarrow 3} \frac{\sqrt{x + 1} - 2}{{x}^{2} - 9}$

$= {\lim}_{x \rightarrow 3} \frac{\sqrt{x + 1} - 2}{{x}^{2} - 9} \times \frac{\sqrt{x + 1} + 2}{\sqrt{x + 1} + 2}$

$= {\lim}_{x \rightarrow 3} \frac{x + 1 - 4}{\left(x + 3\right) \left(x - 3\right) \left(\sqrt{x + 1} + 2\right)}$

$= {\lim}_{x \rightarrow 3} \frac{\cancel{\left(x - 3\right)}}{\left(x + 3\right) \cancel{\left(x - 3\right)} \left(\sqrt{x + 1} + 2\right)}$

$= {\lim}_{x \rightarrow 3} \frac{1}{x + 3} \times \frac{1}{\sqrt{x + 1} + 2}$

$= \left(\frac{1}{6}\right) \left(\frac{1}{4}\right)$

$= \frac{1}{24}$.