Question

Find $\underset{x\to 0}{lim}\frac{\sqrt{a+x}-\sqrt{a}}{x\sqrt{a}(a+x)}$

Answer 1

$\underset{x\to 0}{lim}$ $\frac{\sqrt{a+x}-\sqrt{a}}{x\sqrt{a(a+x)}}=\underset{x\to 0}{lim}$ $\frac{\sqrt{a+x}-\sqrt{a}}{x\sqrt{a(a+x)}}.\frac{\sqrt{a+x}+\sqrt{a}}{\sqrt{a+x}+\sqrt{a}}$

$=\underset{x\to 0}{lim}$ $\frac{x}{x\sqrt{a(a+x)}}.\frac{1}{\sqrt{a+x}+\sqrt{a}}$

$=\frac{1}{\sqrt{a.a}}.\frac{1}{\sqrt{a}+\sqrt{a}}$

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