# How do you find the limit of (sqrt(x+5)-3)/(x-4) as x->4?

Dec 9, 2016

${\lim}_{x \rightarrow 4} \frac{\sqrt{x + 5} - 3}{x - 4} = \frac{1}{6}$

#### Explanation:

${\lim}_{x \rightarrow 4} \frac{\sqrt{x + 5} - 3}{x - 4} = {\lim}_{x \rightarrow 4} \frac{\sqrt{x + 5} - 3}{x - 4} \cdot \frac{\sqrt{x + 5} + 3}{\sqrt{x + 5} + 3}$
$\text{ } = {\lim}_{x \rightarrow 4} \frac{\left(\sqrt{x + 5} - 3\right)}{\left(x - 4\right)} \cdot \frac{\left(\sqrt{x + 5} + 3\right)}{\left(\sqrt{x + 5} + 3\right)}$
$\text{ } = {\lim}_{x \rightarrow 4} \frac{{\sqrt{x + 5}}^{2} - {3}^{2}}{\left(x - 4\right) \left(\sqrt{x + 5} + 3\right)}$
$\text{ } = {\lim}_{x \rightarrow 4} \frac{x + 5 - 9}{\left(x - 4\right) \left(\sqrt{x + 5} + 3\right)}$
$\text{ } = {\lim}_{x \rightarrow 4} \frac{x - 4}{\left(x - 4\right) \left(\sqrt{x + 5} + 3\right)}$
$\text{ } = {\lim}_{x \rightarrow 4} \frac{1}{\sqrt{x + 5} + 3}$
$\text{ } = \frac{1}{\sqrt{4 + 5} + 3}$
$\text{ } = \frac{1}{\sqrt{9} + 3}$
$\text{ } = \frac{1}{3 + 3}$
$\text{ } = \frac{1}{6}$