# How do you find lim (2-sqrt(x+2))/(4-x^2) as x->2 using l'Hospital's Rule?

Jan 8, 2017

${\lim}_{x \to 2} \frac{2 - \sqrt{x + 2}}{4 - {x}^{2}} = \frac{1}{16}$

#### Explanation:

First we observe that:

${\lim}_{x \to 2} 2 - \sqrt{x + 2} = 0$

${\lim}_{x \to 2} 4 - {x}^{2} = 0$

So the limit:

${\lim}_{x \to 2} \frac{2 - \sqrt{x + 2}}{4 - {x}^{2}}$

is in the indeterminate form $\frac{0}{0}$ and we can use l'Hospital's rule:

${\lim}_{x \to 2} \frac{2 - \sqrt{x + 2}}{4 - {x}^{2}} = {\lim}_{x \to 2} \frac{\frac{d}{\mathrm{dx}} \left(2 - \sqrt{x + 2}\right)}{\frac{d}{\mathrm{dx}} \left(4 - {x}^{2}\right)}$

Calculate the derivatives:

$\frac{d}{\mathrm{dx}} \left(2 - \sqrt{x + 2}\right) = - \frac{1}{2 \sqrt{x + 2}}$

$\frac{d}{\mathrm{dx}} \left(4 - {x}^{2}\right) = - 2 x$

So:

${\lim}_{x \to 2} \frac{2 - \sqrt{x + 2}}{4 - {x}^{2}} = {\lim}_{x \to 2} \frac{1}{4 x \left(\sqrt{x + 2}\right)} = \frac{1}{16}$