Evaluate the following limits: _ x 0 ( 1- x x ) - Vidyadip

Question

Evaluate the following limits:$\lim _{x \rightarrow 0}\left(\frac{\sqrt{1-\cos x}}{x}\right)$

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Answer

$\lim _{x \rightarrow 0} \frac{\sqrt{1-\cos x}}{x}$
$=\lim _{x \rightarrow 0} \frac{\sqrt{2 \sin ^2 \frac{x}{2}}}{x}=\lim _{x \rightarrow 0} \frac{\sqrt{2}\left|\sin \frac{x}{2}\right|}{x}$
$\begin{aligned} |x| & =x & & ; x \geq 0 \\ & =-x & & ; x<0 \end{aligned}$
Left hand limit $=\lim _{x \rightarrow 0^{-}} \frac{\sqrt{2}\left|\sin \frac{x}{2}\right|}{x}$
$ =\sqrt{2} \lim _{x \rightarrow 0} \frac{\sin \left(\frac{-x}{2}\right)}{x}$
$=\sqrt{2} \lim _{x \rightarrow 0} \frac{-\sin \frac{x}{2}}{\frac{x}{2}} \times \frac{1}{2}$
$=\sqrt{2} \times(-1) \times \frac{1}{2}$
$\cdots\left[\because \lim _{\theta \rightarrow 0} \frac{\sin p \theta}{p \theta}=1\right]$
$=\frac{-1}{\sqrt{2}}$
Right hand limit $=\lim _{x \rightarrow 0^{+}} \frac{\sqrt{2}\left|\sin \frac{x}{2}\right|}{x}$
$=\sqrt{2} \lim _{x \rightarrow 0} \frac{\sin \left(\frac{x}{2}\right)}{x}$
$=\sqrt{2} \lim _{x \rightarrow 0} \frac{\sin \frac{x}{2}}{\frac{x}{2}} \times \frac{1}{2}$
$=\sqrt{2} \times(1) \times \frac{1}{2}$
$\ldots\left[\because \lim _{\theta \rightarrow 0} \frac{\sin p \theta}{p \theta}=1\right]$
$=\frac{1}{\sqrt{2}}$
$\therefore$ Left hand limit is not equal to the right hand limit
$\therefore \lim _{x \rightarrow 0}\left(\frac{\sqrt{1-\cos x}}{x}\right)$ does not exist.

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