# How do you find the limit of (sqrtx-2)/(x-4) as x->4^-?

Dec 13, 2016

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

#### Explanation:

Since the function,
as $x \to {4}^{-}$, is equal to the indeterminate form: $\frac{0}{0}$

We can apply L'Hopital's Rule:

Therefore:

${\lim}_{x \to {4}^{-}} \frac{\sqrt{x} - 2}{x - 4} \implies \frac{0}{0}$

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