Evaluate

To evaluate:

$\lim _{h \rightarrow 0} \frac{1}{h}\left\{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}\right\}$

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

$\lim _{h \rightarrow 0} \frac{1}{h}\left\{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}\right\}=\frac{0}{0}$

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

$\lim _{h \rightarrow 0} \frac{1}{h}\left\{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}\right\}=\lim _{h \rightarrow 0} \frac{\frac{d}{d h}\left(\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}\right)}{\frac{d}{d h}(h)}$

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

$\lim _{h \rightarrow 0} \frac{1}{h}\left\{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}\right\}=0$

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