Evaluate

To evaluate:

$\lim _{x \rightarrow 1} \frac{\sqrt{3+x}-\sqrt{5-x}}{x^{2}-1}$

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

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

As $x \rightarrow 0$, we have

$\lim _{x \rightarrow 0} \frac{\sqrt{3+x}-\sqrt{5-x}}{x^{2}-1}=\frac{0}{0}$

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

$\lim _{x \rightarrow 1} \frac{\sqrt{3+x}-\sqrt{5-x}}{x^{2}-1}=\lim _{x \rightarrow 1} \frac{\frac{d}{d x}(\sqrt{3+x}-\sqrt{5-x})}{\frac{d}{d x}\left(x^{2}-1\right)}$

$\lim _{x \rightarrow 1} \frac{\sqrt{3+x}-\sqrt{5-x}}{x^{2}-1}=\lim _{x \rightarrow 1} \frac{\frac{1}{2 \sqrt{3+x}}+\frac{1}{2 \sqrt{5-x}}}{2 x}$

$\lim _{x \rightarrow 1} \frac{\sqrt{3+x}-\sqrt{5-x}}{x^{2}-1}=\frac{1}{4}$

Thus, the value of $\lim _{x \rightarrow 1} \frac{\sqrt{3+x}-\sqrt{5-x}}{x^{2}-1}$ is $\frac{1}{4}$