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 $\frac{1}{16}$

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

$\underset{h\to 0}{lt}\frac{\cot (\frac{\pi }{2}+h)-\cos (\frac{\pi }{2}+h)}{{(\pi -2(\frac{\pi }{2}+h))}^3}$

$\Rightarrow \underset{h\to 0}{lt}\frac{-\tan h+\sin h}{-8h^3}$

$\underset{h\to 0}{lt}\frac{-\sin h(1-\cos h)}{\cos h(-8h^3)}$

$\Rightarrow \frac{1}{8}\underset{h\to 0}{lt}\frac{tanh}{h}\times \frac{1-\cos h}{h^2}$

$\frac{1}{8}\underset{h\to 0}{lt}\frac{\tan h}{h}\times \underset{h\to 0}{lt}\frac{2{\sin }^2\frac{h}{2}}{h^2}$

$\Rightarrow \frac{1}{8}\times \frac{2}{4}\underset{h\to 0}{lt}\frac{{\sin }^2\frac{h}{2}}{{\left(\frac{h}{2}\right)}^2}$

$\Rightarrow \frac{1}{16}$.