A) \[\frac{1}{8}\]
B) 0
C) \[\frac{1}{32}\]
D) \[\infty \]
Correct Answer: C
Solution :
\[\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,\frac{\left[ 1-\tan \frac{x}{2} \right][1-\sin x]}{\left[ 1+\tan \frac{x}{2} \right]{{[\pi -2x]}^{3}}}\] Let \[x=\frac{\pi }{2}-h\,as\,x\to \frac{\pi }{2},\,h\to 0\] \[\therefore \,\underset{x\to 0}{\mathop{\lim }}\,\frac{1-\tan \left( \frac{\pi }{4}-\frac{h}{2} \right)}{1+\tan \left( \frac{\pi }{4}-\frac{h}{2} \right)}.\frac{(1-\cos \,h)}{{{(2\,h)}^{3}}}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,\,\,\,\tan \frac{h}{2}.\frac{2{{\sin }^{2}}\frac{h}{2}}{8{{h}^{3}}}\] \[\left[ \because \tan \left( \frac{\pi }{4}-x \right)=\frac{1-\tan x}{1+\tan x} \right]\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,\,\,\frac{1}{4}.\frac{\tan \frac{h}{2}}{\frac{h}{2}\times 2}\times \left( \frac{\sin \frac{h}{2}}{\frac{h}{2}} \right)\times \frac{1}{4}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{1}{32}.\left( \frac{\tan \frac{h}{2}}{\frac{h}{2}} \right)\left( \frac{\sin \frac{h}{2}}{\frac{h}{2}} \right)=\frac{1}{32}\]
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