# Evaluate

Question:

Evaluate

$\lim _{x \rightarrow 0}\left(\frac{\sqrt{2-x}-\sqrt{2+x}}{x}\right)$

Solution:

To evaluate:

$\lim _{x \rightarrow 0} \frac{\sqrt{2-x}-\sqrt{2+x}}{x}$

Formula used: L'Hospital's rule

Let $f(x)$ and $g(x)$ be two functions which are differentiable on an open interval I except at a point a where

$\lim _{x \rightarrow a} \mathrm{f}(\mathrm{x})=\lim _{x \rightarrow a} \mathrm{~g}(\mathrm{x})=0$ or $\pm \infty$

then

$\lim _{x \rightarrow a} \frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}=\lim _{x \rightarrow a} \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{g}^{\prime}(\mathrm{x})}$

As $\mathrm{x} \rightarrow 0$, we have

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

This represents an indeterminate form. Thus applying L'Hospital's rule, we get

$\lim _{x \rightarrow 0} \frac{\sqrt{2-x}-\sqrt{2+x}}{x}=\lim _{x \rightarrow 0} \frac{\frac{d}{d x}(\sqrt{2-x}-\sqrt{2+x})}{\frac{d}{d x}(x)}$

$\lim _{x \rightarrow 0} \frac{\sqrt{2-x}-\sqrt{2+x}}{x}=\lim _{x \rightarrow 0} \frac{-\frac{1}{2 \sqrt{2-x}}-\frac{1}{2 \sqrt{2+x}}}{1}$

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

$\lim _{x \rightarrow 0} \frac{\sqrt{2-x}-\sqrt{2+x}}{x}=\frac{-1}{\sqrt{2}}$

Thus, the value of $\lim _{x \rightarrow 0} \frac{\sqrt{2-x}-\sqrt{2+x}}{x}$ is $\frac{-1}{\sqrt{2}}$