Question

$\underset{x\to \frac{\pi }{2}}{lim}\frac{cotx-cosx}{{(\pi -2x)}^3}$ equals

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

Answer 1

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

Given,

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

apply L-Hospital's rule

$={lim}_{x\to \frac{\pi }{2}}\left(\frac{-{\csc }^2\left(x\right)+\sin \left(x\right)}{-6{(\pi -2x)}^2}\right)$

$={lim}_{x\to \frac{\pi }{2}}(-\frac{-{\csc }^2\left(x\right)+\sin \left(x\right)}{6{(\pi -2x)}^2})$

apply L-Hospital's rule

$={lim}_{x\to \frac{\pi }{2}}\left(\frac{-2{\csc }^2\left(x\right)\cot \left(x\right)-\cos \left(x\right)}{-24(\pi -2x)}\right)$

$={lim}_{x\to \frac{\pi }{2}}(-\frac{-2{\csc }^2\left(x\right)\cot \left(x\right)-\cos \left(x\right)}{24(\pi -2x)})$

apply L-Hospital's rule

$={lim}_{x\to \frac{\pi }{2}}\left(\frac{2(-2{\csc }^2\left(x\right){\cot }^2\left(x\right)-{\csc }^4\left(x\right))-\sin \left(x\right)}{-48}\right)$

$={lim}_{x\to \frac{\pi }{2}}(-\frac{1}{48}(2(-2{\csc }^2\left(x\right){\cot }^2\left(x\right)-{\csc }^4\left(x\right))-\sin \left(x\right)))$

$={lim}_{x\to \frac{\pi }{2}}(-\frac{2(-2{\csc }^2\left(x\right){\cot }^2\left(x\right)-{\csc }^4\left(x\right))-\sin \left(x\right)}{48})$

$=-\frac{2(-2{\csc }^2\left(\frac{\pi }{2}\right){\cot }^2\left(\frac{\pi }{2}\right)-{\csc }^4\left(\frac{\pi }{2}\right))-\sin \left(\frac{\pi }{2}\right)}{48}$

upon solving, we get,

$=\frac{1}{16}$