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Solution:

To evaluate:

$\lim _{x \rightarrow 0} \frac{\sqrt{a+x}-\sqrt{a}}{x \sqrt{a(a+x)}}$

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

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

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

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

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

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

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

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

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