Evaluate

To evaluate:

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

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

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

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

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

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

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

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