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

Let \(\mathrm{L}=\lim _{x \rightarrow \frac{\pi}{2}} \frac{(1-\sin x)\left(8 x^3-\pi^3\right) \cos x}{(\pi-2 x)^4}\) \(=\lim _{x \rightarrow \frac{\pi}{2}} \frac{(1-\sin x)(2 x-\pi)\left(4 x^2+\pi^2+2 \pi x\right) \cos x}{16\left(\frac{\pi}{2}-x\right)^4}\)
Let \(l_1=\lim _{x \rightarrow \frac{\pi}{2}}\left(4 x^2+\pi^2+2 \pi x\right)=3 \pi^2\)
and \(l_2=\lim _{x \rightarrow \frac{\pi}{2}} \frac{(1-\sin x)(2 x-\pi) \cos x}{16\left(\frac{\pi}{2}-x\right)^4}\)
\(\begin{aligned}
& \text { Put } \frac{\pi}{2}-x=h \\
& \Rightarrow x=\frac{\pi}{2}-\mathrm{h} \text { and as } x \rightarrow \frac{\pi}{2}, \mathrm{~h} \rightarrow 0 \\
& \therefore \quad l_2=\lim _{h \rightarrow 0} \frac{1-\sin \left(\frac{\pi}{2}-h\right)(-2 h) \cos \left(\frac{\pi}{2}-h\right)}{16 h^4} \\
& =-\frac{1}{8} \lim _{h \rightarrow 0} \frac{1-\cosh }{h^2} \cdot \frac{\sin h}{h} \\
& =-\frac{1}{8}\left(\frac{1}{2}\right) \cdot 1 \\
& =\frac{-1}{16} \\
& \therefore \quad \mathrm{~L}=3 \pi^2 \times\left(-\frac{1}{16}\right)=\frac{-3 \pi^2}{16}
\end{aligned}\)