lim _{x rightarrow frac{pi}{2}} frac{cot x-co | vedclass

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(B) Let $I = \lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^3}$ $= \lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x(1-\sin x)}{\si

Vedclass · Accepted Answer

$\frac{1}{16}$

Vedclass · Answer

(B) Let $I = \lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^3}$ $= \lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x(1-\sin x)}{\sin x(\pi-2 x)^3}$ Substitute $x = \frac{\pi}{2}-h$,then $\pi-2x = 2h$. As $x \rightarrow \frac{\pi}{2}$,$h \rightarrow 0$. $I = \lim _{h \rightarrow 0} \frac{\cos(\frac{\pi}{2}-h)(1-\sin(\frac{\pi}{2}-h))}{\sin(\frac{\pi}{2}-h)(2h)^3}$ $= \lim _{h \rightarrow 0} \frac{\sin h(1-\cos h)}{\cos h \cdot 8h^3}$ Using $1-\cos h = 2\sin^2(\frac{h}{2})$,$I = \lim _{h \rightarrow 0} \frac{\sin h \cdot 2\sin^2(\frac{h}{2})}{\cos h \cdot 8h^3} = \frac{2}{8} \lim _{h \rightarrow 0} \frac{\sin h}{h} \cdot \lim _{h \rightarrow 0} \frac{\sin^2(\frac{h}{2})}{(\frac{h}{2})^2 \cdot 4} \cdot \lim _{h \rightarrow 0} \frac{1}{\cos h}$ $= \frac{1}{4} \cdot 1 \cdot (1)^2 \cdot \frac{1}{4} \cdot 1 = \frac{1}{16}$

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^3}$ equals

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