Question

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

• A. $\frac{1}{24}$
• B. $\frac{1}{16}$
• C. $\frac{1}{81}$
• D. $\frac{1}{4}$

Answer 1

Answer: (B)

$\frac{1}{16}$

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

Taking , $(\frac{\pi }{2}-x)=t$

We get: $\underset{x\to \frac{\pi }{2}}{lim}\frac{\cos x}{\sin x}(1-\sin x)\times \frac{1}{t^3}$

Taking $\cot x=\frac{\cos x}{\sin x}$ for the above equation adn taking $\cos x$ common.

Now,

$(\sin (\frac{\pi }{2}-t)=\cos t)$

Hence, we get $\underset{x\to \frac{\pi }{2}}{lim}\frac{\sin t}{\cos t}(1-\cos t)\times \frac{1}{t^3}$

$\underset{x\to \frac{\pi }{2}}{lim}\frac{\sin t\left(2{\sin }^2\frac{t}{2}\right)}{\cos t}$

Using L'Hospital rule

$\underset{x\to \frac{\pi }{2}}{lim}\frac{1}{2^2}\times \frac{\sin t}{t}\times \frac{2{\sin }^2\frac{t}{2}}{4{\left(\frac{t}{2}\right)}^2}\times \frac{1}{cost}$

Hence we get: $\frac{1}{2\times 8}=\frac{1}{16}$