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$\lim _{x \rightarrow \pi / 2} \frac{\left(1-\tan \left(\frac{x}{2}\right)\right)(1-\sin x)}{\left(1+\tan \left(\frac{x}{2}\right)\right)(\pi-2 x)^{3}}=?$
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Verified Answer
The correct answer is:
$1 / 32$
Put $x=\frac{\pi}{2}-h$ as $x \rightarrow \frac{\pi}{2}, h \rightarrow 0$
$\therefore$ Given limit
$=\lim _{h \rightarrow 0} \frac{1-\tan \left(\frac{\pi}{4}-\frac{\mathrm{h}}{2}\right)}{1+\tan \left(\frac{\pi}{4}-\frac{\mathrm{h}}{2}\right)} \cdot \frac{(1-\cosh )}{(2 \mathrm{~h})^{3}}$
$=\lim _{\mathrm{h} \rightarrow 0} \tan \frac{\mathrm{h}}{2} \frac{2 \sin ^{2} \frac{\mathrm{h}}{2}}{8 \mathrm{~h}^{3}}$
$=\lim _{\mathrm{h} \rightarrow 0} \frac{1}{4} \cdot \frac{\tan \frac{\mathrm{h}}{2}}{\frac{\mathrm{h}}{2} \times 2}\left(\frac{\sin \frac{\mathrm{h}}{2}}{\frac{\mathrm{h}}{2}}\right)^{2} \times \frac{1}{4}$
$=\lim _{\mathrm{h} \rightarrow 0} \frac{1}{32} \cdot\left(\frac{\tan \frac{\mathrm{h}}{2}}{\frac{\mathrm{h}}{2} \times 2}\right)\left(\frac{\sin \frac{\mathrm{h}}{2}}{\frac{\mathrm{h}}{2}}\right)^{2}=\frac{1}{32}$
$\therefore$ Given limit
$=\lim _{h \rightarrow 0} \frac{1-\tan \left(\frac{\pi}{4}-\frac{\mathrm{h}}{2}\right)}{1+\tan \left(\frac{\pi}{4}-\frac{\mathrm{h}}{2}\right)} \cdot \frac{(1-\cosh )}{(2 \mathrm{~h})^{3}}$
$=\lim _{\mathrm{h} \rightarrow 0} \tan \frac{\mathrm{h}}{2} \frac{2 \sin ^{2} \frac{\mathrm{h}}{2}}{8 \mathrm{~h}^{3}}$
$=\lim _{\mathrm{h} \rightarrow 0} \frac{1}{4} \cdot \frac{\tan \frac{\mathrm{h}}{2}}{\frac{\mathrm{h}}{2} \times 2}\left(\frac{\sin \frac{\mathrm{h}}{2}}{\frac{\mathrm{h}}{2}}\right)^{2} \times \frac{1}{4}$
$=\lim _{\mathrm{h} \rightarrow 0} \frac{1}{32} \cdot\left(\frac{\tan \frac{\mathrm{h}}{2}}{\frac{\mathrm{h}}{2} \times 2}\right)\left(\frac{\sin \frac{\mathrm{h}}{2}}{\frac{\mathrm{h}}{2}}\right)^{2}=\frac{1}{32}$
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