MCQ
$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\cot x - \cos x}}{{{{\left( {\pi - 2x} \right)}^3}}} = $ . . . .
- A$\frac{1}{4}$
- B$\frac{1}{{24}}$
- C$\frac{1}{{16}}$
- D$\frac{1}{8}$
Put $\frac{\pi }{2} - x = t \Rightarrow $ as $x \to \frac{\pi }{2} \Rightarrow t \to 0$
$ = \mathop {\lim }\limits_{t \to 0} \frac{{\cot \left( {\frac{\pi }{2} - t} \right)\left( {1 - \sin \left( {\frac{\pi }{2} - t} \right)} \right)}}{{8{t^3}}}$
$ = \mathop {\lim }\limits_{t \to 0} \frac{{\tan \,t\left( {1 - \cos \,t} \right)}}{{8{t^3}}}$
$ = \mathop {\lim }\limits_{t \to 0} \frac{{\tan \,t}}{{8t}}.\frac{{1 - \cos t}}{{{t^2}}}$
$ = \frac{1}{8}.1.\frac{1}{2} = \frac{1}{{16}}$
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