lim _{x rightarrow frac{pi}{2}} frac{(1-sin x | vedclass

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(C) Let $L = \lim _{x \rightarrow \frac{\pi}{2}} \frac{(1-\sin x)(8 x^3-\pi^3) \cos x}{(\pi-2 x)^4}$.We can rewrite the expression as:$L = \lim _{x \r

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$\frac{-3 \pi^2}{16}$

Vedclass · Answer

(C) Let $L = \lim _{x \rightarrow \frac{\pi}{2}} \frac{(1-\sin x)(8 x^3-\pi^3) \cos x}{(\pi-2 x)^4}$.We can rewrite the expression as:$L = \lim _{x \rightarrow \frac{\pi}{2}} \frac{(1-\sin x)(2x-\pi)(4x^2+\pi^2+2\pi x) \cos x}{16(\frac{\pi}{2}-x)^4}$.Let $x - \frac{\pi}{2} = h$,so $x = \frac{\pi}{2} + h$. As $x \rightarrow \frac{\pi}{2}$,$h \rightarrow 0$.Then $\cos x = \cos(\frac{\pi}{2} + h) = -\sin h$ and $1 - \sin x = 1 - \sin(\frac{\pi}{2} + h) = 1 - \cos h$.Also,$2x - \pi = 2(\frac{\pi}{2} + h) - \pi = 2h$.Substituting these into the limit:$L = \lim _{h}$ ${\rightarrow 0} \frac{(1-\cos h)(2h)(4(\frac{\pi}{2}+h)^2 + \pi^2 + 2\pi(\frac{\pi}{2}+h))(-\sin h)}{16(-h)^4}$.$L = \lim _{h \rightarrow 0} \frac{(1-\cos h)(2h)(3\pi^2 + 8\pi h + 4h^2)(-\sin h)}{16h^4}$.$L = \lim _{h \rightarrow 0} \frac{-(1-\cos h)}{h^2} \cdot \frac{\sin h}{h} \cdot \frac{2h(3\pi^2 + 8\pi h + 4h^2)}{16h}$.$L = -(\frac{1}{2}) \cdot (1) \cdot \frac{2(3\pi^2)}{16} = -\frac{1}{2} \cdot \frac{6\pi^2}{16} = -\frac{3\pi^2}{16}$.

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{(1-\sin x)(8 x^3-\pi^3) \cos x}{(\pi-2 x)^4}$

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