Answer
$$\lim _{h \rightarrow 0} \frac{\sqrt{a+2 h}-\sqrt{a}}{h} =\frac{1 }{ \sqrt{a}}$$
Work Step by Step
Given $$\lim _{h \rightarrow 0} \frac{\sqrt{a+2 h}-\sqrt{a}}{h} $$
let $$ f(h) = \frac{\sqrt{a+2 h}-\sqrt{a}}{h} $$
Since, we have
$$ f(0)=\frac{\sqrt{a}-\sqrt{a}}{0}=\frac{0}{0}$$
So, transform algebraically and cancel
\begin{aligned}L&= \lim _{h \rightarrow 0} \frac{\sqrt{a+2 h}-\sqrt{a}}{h} \\
&= \lim _{h \rightarrow 0} \frac{\sqrt{a+2 h}-\sqrt{a}}{h}\times \frac{\sqrt{a+2 h}+\sqrt{a}}{\sqrt{a+2 h}+\sqrt{a}} \\ \\
&=\lim _{h \rightarrow 0} \frac{(\sqrt{a+2 h})^{2}-(\sqrt{a})^{2}}{h(\sqrt{a+2 h}+\sqrt{a})} \\ &=\lim _{h \rightarrow 0} \frac{a+2 h-a}{h(\sqrt{a+2 h}+\sqrt{a})} \\ &=\lim _{h \rightarrow 0} \frac{2 h}{h(\sqrt{a+2 h}+\sqrt{a})}\\
&=\lim _{h \rightarrow 0} \frac{2 }{ (\sqrt{a+2 h}+\sqrt{a})} \\
&=\frac{2 }{ (\sqrt{a+0}+\sqrt{a})}\\
&=\frac{2 }{ 2\sqrt{a}}\\
&=\frac{1 }{ \sqrt{a}}
\end{aligned}
