Question
Evaluate the following limits:
$\lim _{x \rightarrow a}\left[\frac{\sin (\sqrt{x})-\sin (\sqrt{a})}{x-a}\right]$

Answer

$\lim _{x \rightarrow \mathrm{a}}\left[\frac{\sin (\sqrt{x})-\sin (\sqrt{\mathrm{a}})}{x-\mathrm{a}}\right]$
Put $\sqrt{x}=y, \sqrt{\mathrm{a}}=\mathrm{b}$ and $y=\mathrm{b}+\mathrm{h}$.
As $x \rightarrow \mathrm{a}, y \rightarrow \mathrm{b}$ and $\mathrm{h} \rightarrow 0$.
$ \lim _{x \rightarrow a} \frac{\sin \sqrt{x}-\sin \sqrt{a}}{x-a}$
$=\lim _{y \rightarrow b} \frac{\sin y-\sin b}{y^2-b^2}$
$=\lim _{y \rightarrow b} \frac{\sin y-\sin b}{(y-b)(y+b)}$
$=\lim _{h \rightarrow 0} \frac{\sin (b+h)-\sin b}{h(b+h+b)} $
$=\lim _{h \rightarrow 0} \frac{2 \cos \left(\frac{b+h+b}{2}\right) \sin \left(\frac{b+h-b}{2}\right)}{h(2 b+h)}$
$=\lim _{h \rightarrow 0} \frac{2 \cos \left(b+\frac{h}{2}\right) \cdot \sin \left(\frac{h}{2}\right)}{h(2 b+h)}$
$=\lim _{h \rightarrow 0} \frac{\cos \left[b+\frac{h}{2}\right]}{2 b+h} \cdot \frac{\sin \frac{h}{2}}{\frac{h}{2}}$
$=\lim _{h \rightarrow 0} \frac{\cos \left[b+\frac{h}{2}\right]}{2 b+h} \cdot\left[\lim _{h \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}}\right]$
$=\frac{\cos (b+0)}{2 b+0} \cdot 1$
$=\frac{\cos b}{2 b}$
$=\frac{\cos \sqrt{a}}{2 \sqrt{a}} $

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