Answer
$$\lim _{x \rightarrow a} \frac{\sqrt{x}-\sqrt{a}}{x-a} =\frac{1}{2\sqrt{a}}$$
Work Step by Step
Given $$\lim _{x \rightarrow a} \frac{\sqrt{x}-\sqrt{a}}{x-a},\ \ \ a \ \ \text{is constant}$$
let $$ f(x) = \frac{\sqrt{x}-\sqrt{a}}{x-a}$$
Since, we have
$$ f(a)= \frac{\sqrt{a}-\sqrt{a}}{a-a}=\frac{0}{0}$$
So, we get
\begin{aligned}
L&=\lim _{x \rightarrow a} \frac{\sqrt{x}-\sqrt{a}}{x-a}\\
&=\lim _{x \rightarrow a} \frac{\sqrt{x}-\sqrt{a}}{x-a} \times \frac{\sqrt{x}+\sqrt{a}}{\sqrt{x}+\sqrt{a}}\\
&=\lim _{x \rightarrow a} \frac{(\sqrt{x})^{2}-(\sqrt{a})^{2}}{(x-a)(\sqrt{x}+\sqrt{a})}\\
&=\lim _{x \rightarrow a} \frac{x-a}{(x-a)(\sqrt{x}+\sqrt{a})}\\
&=\lim _{x \rightarrow a} \frac{1}{(\sqrt{x}+\sqrt{a})} \\
&=\frac{1}{(\sqrt{a}+\sqrt{a})} \\
&=\frac{1}{2\sqrt{a}}
\end{aligned}
