Evaluate the following limits: _ x 1 [1-x^2 x ] - Vidyadip

Question

Evaluate the following limits:
$\lim _{x \rightarrow 1}\left[\frac{1-x^2}{\sin \pi x}\right]$

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Answer

$\lim _{x \rightarrow 1} \frac{1-x^2}{\sin \pi x}=\lim _{x \rightarrow 1} \frac{(1-x)(1+x)}{\sin \pi x}$
Put $1-x=\mathrm{h}$,
$\therefore \quad x=1-\mathrm{h}$
$\text { As } x \rightarrow 1, \mathrm{~h} \rightarrow 0$
$
\begin{aligned}
\therefore \quad & \lim _{x \rightarrow 1} \frac{(1-x)(1+x)}{\sin \pi x} \\
= & \lim _{h \rightarrow 0} \frac{(h)(1+1-h)}{\sin \pi(1-h)} \\
= & \lim _{h \rightarrow 0} \frac{(h)(2-h)}{\sin (\pi-\pi h)} \\
& =\lim _{h \rightarrow 0} \frac{(h)(2-h)}{\sin \pi h} \quad \ldots[\because \sin (\pi-\theta)=\sin \theta]
\end{aligned}
$
$ =\lim _{h \rightarrow 0} \frac{(2-h)^{}}{\left(\frac{\sin \pi h}{\pi h}\right) \cdot \pi}$
$=\frac{1}{\pi} \cdot \frac{\lim _{h \rightarrow 0}(2-h)}{\lim _{h \rightarrow 0}\left(\frac{\sin \pi h}{\pi h}\right)}$
$=\frac{1}{\pi} \cdot \frac{(2-0)}{1} \quad \ldots\left[\because \lim _{\theta \rightarrow 0} \frac{\sin p \theta}{p \theta}=1\right] $
$=\frac{2}{\pi}$

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