Question

The value of $\underset{x\to \left(\frac{\pi }{2}\right)}{lim}\frac{[1-\tan \left(\frac{x}{2}\right)][1-\sin x]}{[1+\tan \left(\frac{x}{2}\right)]{[\pi -2x]}^3}$ is

• A. $0$
• B. $\infty$
• C. $\frac{1}{32}$
• D. $\frac{1}{8}$

Answer 1

Answer: (C)

$\frac{1}{32}$

$\underset{x\to \left(\frac{\pi }{2}\right)}{lim}\frac{[1-\tan \left(\frac{x}{2}\right)][1-\sin x]}{[1+\tan \left(\frac{x}{2}\right)]{[\pi -2x]}^3}$

$=\underset{x\to \left(\frac{\pi }{2}\right)}{lim}\frac{\tan \left(\frac{\pi }{4}\right)-\tan \left(\frac{x}{2}\right)}{1+\tan \left(\frac{x}{4}\right)\times \tan \left(\frac{x}{2}\right)}\times \frac{(1-\sin x)}{{(\pi -2x)}^3}$

$=\underset{x\to \left(\frac{\pi }{2}\right)}{lim}\tan (\frac{\pi }{4}-\frac{x}{2})\times \frac{(1-\sin x)}{{(\pi -2x)}^3}$

Substituting $x=(\frac{\pi }{2}-h)$ in the expression, we get

$=\underset{h\to 0}{lim}\tan [\frac{\pi }{4}-\frac{1}{2}(\frac{\pi }{2}-h)].\frac{1-\sin (\frac{\pi }{2}-h)}{{[\pi -2(\frac{\pi }{2}-h)]}^3}$

$=\underset{h\to 0}{lim}\tan (\frac{\pi }{4}-\frac{\pi }{4}+\frac{h}{2})\frac{1-\sin (\frac{\pi }{2}-h)}{{[\pi -\pi +2h]}^3}$

$=\underset{h\to 0}{lim}\tan \left(\frac{h}{2}\right).\frac{1-\cos h}{{\left(2h\right)}^3}=\underset{h\to 0}{lim}\tan \left(\frac{h}{2}\right)\frac{[1-1+2{\sin }^2\left(\frac{h}{2}\right)]}{8h^3}$

$=\underset{h\to 0}{lim}\frac{\tan \left(\frac{h}{2}\right)\times 2\times {\sin }^2\left(\frac{h}{2}\right)}{2\times \left(\frac{h}{2}\right)\times 8\times 4{\left(\frac{h}{2}\right)}^2}=\frac{1}{32}\underset{h\to 0}{lim}\frac{\tan \left(\frac{h}{2}\right)}{\left(\frac{h}{2}\right)}\times {\left(\frac{\sin \left(\frac{h}{2}\right)}{\left(\frac{h}{2}\right)}\right)}^2$

$=\frac{1}{32}\times 1\times 1=\frac{1}{32}$