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

Question

Aicharya · Accepted Answer

Let $\mathrm{L}=\lim _{x ightarrow \frac{\pi}{2}} \frac{(1-\sin x)\left(8 x^3-\pi^3ight) \cos x}{(\pi-2 x)^4}$ $=\lim _{x ightarrow \frac{\pi}{2}} \frac{(1-\sin x)(2 x-\pi)\left(4 x^2+\pi^2+2 \pi xight) \cos x}{16\left(\frac{\pi}{2}-xight)^4}$ Let $l_1=\lim _{x ightarrow \frac{\pi}{2}}\left(4 x^2+\pi^2+2 \pi xight)=3 \pi^2$ and $l_2=\lim _{x ightarrow \frac{\pi}{2}} \frac{(1-\sin x)(2 x-\pi) \cos x}{16\left(\frac{\pi}{2}-xight)^4}$ $\begin{aligned} & 	ext { Put } \frac{\pi}{2}-x=h \ & \Rightarrow x=\frac{\pi}{2}-\mathrm{h} 	ext { and as } x ightarrow \frac{\pi}{2}, \mathrm{~h} ightarrow 0 \ & 	herefore \quad l_2=\lim _{h ightarrow 0} \frac{1-\sin \left(\frac{\pi}{2}-hight)(-2 h) \cos \left(\frac{\pi}{2}-hight)}{16 h^4} \ & =-\frac{1}{8} \lim _{h ightarrow 0} \frac{1-\cosh }{h^2} \cdot \frac{\sin h}{h} \ & =-\frac{1}{8}\left(\frac{1}{2}ight) \cdot 1 \ & =\frac{-1}{16} \ & 	herefore \quad \mathrm{~L}=3 \pi^2 	imes\left(-\frac{1}{16}ight)=\frac{-3 \pi^2}{16} \end{aligned}$

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

Solution