Question

${lim}_{x\to \frac{\pi }{2}}\left(\frac{\cos x}{(\pi -2x)}\right)$ equal to ??

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

Answer 1

The correct option is D $\frac{1}{2}$

We have,

${lim}_{x\to \frac{\pi }{2}}\left(\frac{\cos x}{(\pi -2x)}\right)$

This is the $\frac{0}{0}$ form.

So, apply L-Hospital rule

${lim}_{x\to \frac{\pi }{2}}\left(\frac{0-\sin x}{(0-2)}\right)$

${lim}_{x\to \frac{\pi }{2}}\left(\frac{\sin x}{2}\right)$

$=\frac{\sin \frac{\pi }{2}}{2}$

$=\frac{1}{2}$

Hence, this is the answer.