# What is the limit of (sqrt(x+5) - 3)/(x - 4) as x approaches 4?

May 18, 2015

Let's multiply both numerator and denominator of this expression by $\sqrt{x + 5} + 3$ to get rid of undefined $\frac{0}{0}$ value.

Assuming $x \ne 4$ in our expression,

$\frac{\sqrt{x + 5} - 3}{x - 4} =$

$= \frac{\left(\sqrt{x + 5} - 3\right) \cdot \left(\sqrt{x + 5} + 3\right)}{\left(x - 4\right) \cdot \left(\sqrt{x + 5} + 3\right)} =$

$= \frac{x + 5 - 9}{\left(x - 4\right) \cdot \left(\sqrt{x + 5} + 3\right)} =$

$= \frac{x - 4}{\left(x - 4\right) \cdot \left(\sqrt{x + 5} + 3\right)} =$

$= \frac{1}{\sqrt{x + 5} + 3}$

As $x \to 4$, this expression tends to $\frac{1}{\sqrt{4 + 5} + 3} = \frac{1}{6}$.

Therefore,

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

Nov 1, 2015

$\frac{1}{6}$

#### Explanation:

${\lim}_{x \to 4} \left(\sqrt{x + 5} - 3\right) = 0$

${\lim}_{x \to 4} \left(x - 4\right) = 0$

Using the L'Hospital Rule,

${\lim}_{x \to 4} \frac{\sqrt{x + 5} - 3}{x - 4}$

$= {\lim}_{x \to 4} \frac{\frac{d}{\mathrm{dx}} \left(\sqrt{x + 5} - 3\right)}{\frac{d}{\mathrm{dx}} \left(x - 4\right)}$

$= {\lim}_{x \to 4} \frac{\left(\frac{1}{2 \sqrt{x + 5}}\right)}{1}$

$= \frac{1}{2 \sqrt{4 + 5}}$

$= \frac{1}{6}$