mathop {lim }limits_{x to frac{pi }{2}} frac{ | vedclass

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Question

(C) Let $L = \mathop {\lim }\limits_{x 	o \frac{\pi }{2}} \frac{{\cot x - \cos x}}{{{{\left( {\pi - 2x} ight)}^3}}}$.We can rewrite the expression

Vedclass · Accepted Answer

$\frac{1}{16}$

Vedclass · Answer

(C) Let $L = \mathop {\lim }\limits_{x 	o \frac{\pi }{2}} \frac{{\cot x - \cos x}}{{{{\left( {\pi - 2x} ight)}^3}}}$.We can rewrite the expression as:$L = \mathop {\lim }\limits_{x 	o \frac{\pi }{2}} \frac{{\cos x \left( {\frac{1}{{\sin x}} - 1} ight)}}{{ - 8{{\left( {x - \frac{\pi }{2}} ight)}^3}}} = \mathop {\lim }\limits_{x 	o \frac{\pi }{2}} \frac{{\cos x(1 - \sin x)}}{{\sin x \cdot 8{{\left( {\frac{\pi }{2} - x} ight)}^3}}}$.Substitute $t = \frac{\pi }{2} - x$. As $x 	o \frac{\pi }{2}$,$t 	o 0$. Then $x = \frac{\pi }{2} - t$ and $\cos x = \sin t$,$\sin x = \cos t$.$L = \mathop {\lim }\limits_{t 	o 0} \frac{{\sin t(1 - \cos t)}}{{8{t^3}\cos t}}$.$L = \frac{1}{8} \cdot \mathop {\lim }\limits_{t 	o 0} \left( \frac{{\sin t}}{t} ight) \cdot \left( \frac{{1 - \cos t}}{{{t^2}}} ight) \cdot \frac{1}{{\cos t}}$.Using standard limits $\mathop {\lim }\limits_{t 	o 0} \frac{{\sin t}}{t} = 1$ and $\mathop {\lim }\limits_{t 	o 0} \frac{{1 - \cos t}}{{{t^2}}} = \frac{1}{2}$:$L = \frac{1}{8} \cdot 1 \cdot \frac{1}{2} \cdot 1 = \frac{1}{{16}}$.

$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\cot x - \cos x}}{{{{\left( {\pi - 2x} \right)}^3}}} = $ . . . .

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