<s>[INST] How do you evaluate limit of $\frac{\sqrt{a+x} - \sqrt a}{x\sqrt a(\sqrt{a+x})}$ when $x$ tends to $0$? I tried rationalization method where we multiply both the numerator and denominator with appropriate opposite factor of numerator. But I could only get $\frac{1}{\sqrt {a^2+ax}(2 \sqrt {a + \sqrt x)}}$. But the final solution in textbook says it should be $\frac{1}{2a^{3/2}}$. Please help. 
 [/INST] Easier solution: Write $y=\sqrt{x+a}$ and $b=\sqrt{a}$, then
\begin{align}\lim_{x \to 0}\frac{\sqrt{a+x}-\sqrt{a}}{x\sqrt{a}\sqrt{a+x}} &= \lim_{y \to b}\frac{y-b}{by(y^2-b^2)}\\&=\lim_{y \to b}\frac{1}{by(y+b)}\\
&= \frac{1}{2b^3}=\frac{1}{2a\sqrt{a}}\end{align}
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