<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] \begin{align}\lim_{x \to 0}\frac{\sqrt{a+x}-\sqrt{a}}{x\sqrt{a}\sqrt{a+x}} &= \lim_{x \to 0}\frac{a+x-a}{x\sqrt{a}\sqrt{a+x}(\sqrt{a+x}+\sqrt{a})}\\&=\lim_{x\to 0}\frac{1}{\sqrt{a}\sqrt{a+x}(\sqrt{a+x}+\sqrt{a})}\\
&= \frac{1}{\sqrt{a}\sqrt{a}(2\sqrt{a})}\\
&=\frac{1}{2a^\frac32}\end{align}
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