Question

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

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

Answer 1

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

$\underset{x\to \frac{\pi }{2}}{lim}\frac{\cot x-\cos x}{(\pi -2x{)}^3}$
Put $\pi -2x=t\Rightarrow x=\frac{\pi }{2}-\frac{t}{2}$
as $x\to \frac{\pi }{2}$ then $t\to 0$
$=\underset{t\to 0}{lim}\frac{\cot \left(\frac{\pi }{2}-\frac{t}{2}\right)-\cos \left(\frac{\pi }{2}-\frac{t}{2}\right)}{t^3}$
$=\underset{t\to 0}{lim}\frac{\tan \frac{t}{2}-\sin \frac{t}{2}}{t^3}$
$=\underset{t\to 0}{lim}\frac{\frac{\sin \frac{t}{2}}{\cos \frac{t}{2}}-\sin \frac{t}{2}}{t^3}$
$=\underset{t\to 0}{lim}\frac{\sin \frac{t}{2}(1-\cos \frac{t}{2})}{\cos \frac{t}{2}{t}^3}$
$=\underset{t\to 0}{lim}\frac{\sin \frac{t}{2}\times \left({\sin }^2\frac{t}{4}\right)}{\cos \frac{t}{2}\times \frac{t}{2}\times 2\times \frac{t^2}{16}\times 16}$
$=\underset{t\to 0}{lim}\frac{1}{\cos \frac{t}{2}}\times \frac{\sin \frac{t}{2}}{\frac{t}{2}}\times \frac{{\sin }^2\frac{t}{4}}{{\left(\frac{t}{4}\right)}^2}\times \frac{2}{16\times 2}$
$=\frac{1}{16}$