Question

$\underset{x\to \pi /2}{lim}\frac{\cot x-\cos x}{{\left(\pi -2x\right)}^3}$ equals

• A. $\frac{1}{24}$
• B. $\frac{1}{16}$
• C. $\frac{1}{8}$
• D. $\frac{1}{4}$

Answer 1

The correct option is B $\frac{1}{16}$

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

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

$=\frac{1}{8}{lim}_{h\to 0}\frac{\cos (\frac{\pi }{2}-h)[1-\sin (\frac{\pi }{2}-h)]}{\sin (\frac{\pi }{2}-h){(\frac{\pi }{2}-\frac{\pi }{2}+h)}^3}$

$=\frac{1}{8}{lim}_{h\to 0}\frac{\sin h(1-\cos h)}{\cos h{h}^3}$

$=\frac{1}{8}{lim}_{h\to 0}\frac{\sin h{\sin }^2\frac{h}{2}}{\cos h{h}^3}$

$=\frac{1}{8}{lim}_{h\to 0}\left(\frac{\sin h}{h}\right){\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)}^2\frac{1}{\cos h}.\frac{1}{4}$

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