Discussion Forum

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

Answer:

Explanation:

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

      =  $\lim_{x \rightarrow \frac{\pi}{2}}\frac{1}{8} \frac{\cos x(1-\sin x)}{\sin x(\frac{\pi}{2}-x)^{3}}$

  = $\lim_{h \rightarrow 0}\frac{1}{8} \frac{\cos (\frac{\pi}{2}-h)(1-\sin (\frac{\pi}{2}-h))}{\sin(\frac{\pi}{2}-h) (\frac{\pi}{2}-\frac{\pi}{2}+h)^{3}}$

    =  $\frac{1}{8}\lim_{h \rightarrow 0}\frac{\sin h(1-\cosh)}{\cosh.h^{3}}$

 = $\frac{1}{8}\lim_{h \rightarrow 0}\frac{\sin h(2\sin^{2}\frac{h}{2})}{\cosh.h^{3}}$

  = $\frac{1}{4}\lim_{h \rightarrow 0}\frac{\sin h.\sin^{2}(\frac{h}{2})}{\cosh.h^{3}}$

=    $\frac{1}{4}\lim_{h \rightarrow 0}(\frac{\sinh}{h})(\frac{\sin\frac{h}{2}}{\frac{h}{2}})^{2}.\frac{1}{\cosh}.\frac{1}{4}$

=   $\frac{1}{4}\times\frac{1}{4}=\frac{1}{16}$