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  4. lim limitsx→0 (sin2x/√2 -√1 + cos x) equal to a √2 , then a is

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Q. $\lim\limits_{x\to0} \frac{sin^{2}x}{\sqrt{2} -\sqrt{1 + cos\, x}}$ equal to $a \sqrt{2}$ , then $a$ is

Limits and Derivatives

Solution:

$\lim\limits_{x\to0} \frac{sin^{2}x}{\sqrt{2} -\sqrt{1 + cos\, x}}$
$= \lim\limits_{x\to0} \frac{sin^2 x }{\sqrt{2} - \sqrt{2\,cos^2 \frac{x}{2}}} \,\,[\frac{0}{0}]$
$= \lim\limits_{x\to0} \frac{sin^2\,x}{\sqrt{2}[ 1 - cos \frac{x}{2}]} = \lim\limits_{x\to0} \frac{sin^2\,x}{2\sqrt{2}\, sin^2 \frac{x}{4}}$
$=\lim\limits_{x\to0} \frac{\left(\frac{sin\,x}{ x}\right)^{2} . 16}{2\sqrt{2}\left(\frac{sin \frac{x}{4}}{\frac{x}{4}}\right)^{2}}$
$ = \frac{16}{2\sqrt{2}} = 4\sqrt{2}$
$\Rightarrow a \sqrt{2} = 4 \sqrt{2}$
$\Rightarrow a = 4$