Question

$\underset{x\to 0}{\lim }\frac{\sqrt{a+x}-\sqrt{a}}{x\sqrt{a^2+ax}}$

Answer 1

$\underset{x\to 0}{\lim }\left[\frac{\sqrt{a+x}-\sqrt{a}}{x\sqrt{a^2+ax}}\right]$

It is of the form $\frac{0}{0}$.
Rationalising the numerator:

$\underset{x\to 0}{\lim }\left[\frac{\left(\sqrt{a+x}-\sqrt{a}\right)\times \left(\sqrt{a+x}+\sqrt{a}\right)}{\left(x\sqrt{a^2+ax}\right)\times \left(\sqrt{a+x}+\sqrt{a}\right)}\right]$

= $\underset{x\to 0}{\lim }\left[\frac{a+x-a}{x\left(\sqrt{a^2+ax}\right)\left(\sqrt{a+x}+\sqrt{a}\right)}\right]$

= $\frac{1}{\left(\sqrt{a^2}\right)\left(\sqrt{a}+\sqrt{a}\right)}$

= $\frac{1}{2a\sqrt{a}}$