Question

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

• A. $\frac{1}{16}$
• B. $\frac{1}{8}$
• C. $\frac{1}{4}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (A)

$\frac{1}{16}$

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

$={lim}_{x\to \pi /2}\frac{\tan (\left(\pi /2\right)-x)[1-\cos (\left(\pi /2\right)-x)]}{{(\pi -2x)}^3}$

${lim}_{x\to \pi /2}\frac{\tan (\left(\pi /2\right)-x)\left[2{\sin }^2\left(\right(\pi /4)-(x/2\left)\right)\right]}{2\left(\right(\pi /2)-x).16{\left(\right(\pi /4)-(x/2\left)\right)}^2}=\frac{2}{32}=\frac{1}{16}$