# How do you find lim (sqrt(9-x)-3)/x as x->0 using l'Hospital's Rule?

Dec 25, 2016

The answer is $= - \frac{1}{6}$

#### Explanation:

We need

$\left(\sqrt{x}\right) ' = \frac{1}{2 \sqrt{x}}$

${\lim}_{x \to 0} \frac{\sqrt{9 - x} - 3}{x} = \frac{0}{0}$

This is undefined, so we apply L'Hôspital's Rule

${\lim}_{x \to 0} \frac{\sqrt{9 - x} - 3}{x} = {\lim}_{x \to 0} \frac{\left(\sqrt{9 - x} - 3\right) '}{x '}$

$= {\lim}_{x \to 0} \left(- \frac{1}{2 \sqrt{9 - x} - 3}\right)$

$= - \frac{1}{6}$