Question

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

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

Answer 1

Answer: (C)

$\frac{1}{32}$

$\underset{x\to \frac{\pi }{2}}{lim}$ $\left(\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}\right)\left(\frac{1-sinx}{{(\pi -2x)}^3}\right)$ $=$ $\underset{x\to \frac{\pi }{2}}{lim}$ $\left(\frac{\tan \pi /4-tanx/2}{1+\tan \pi /4.tanx/2}\right)\left(\frac{1-\cos (\frac{\pi }{2}-x)}{{(\pi -2x)}^3}\right)$

We know that, $(tan\pi /4=1)$ & $(cos(\pi /2-\theta )=sin\theta )$

$\frac{tanA-tanB}{1+tanA.tanB}=tanA-tanB$

and $1-\cos 2\theta =2{\sin }^2\theta$

Using the above identities we can further reduce

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

$=\underset{x\to \pi /2}{lim}$ $\frac{\tan (\pi /4-x/2)}{64(\pi /4-x/2)}\times \frac{2{\sin }^2(\pi /4-x/2)}{{(\pi /4-x/2)}^2}$

$=\frac{1}{32}\underset{x\to \pi /2}{lim}$ $\frac{\tan (\pi /4-x/2)}{(\pi /4-x/2)}\times \frac{{\sin }^2(\pi /4-x/2)}{(\pi /4-x/2)}$

Again, We know that,

$\underset{\theta \to 0}{lim}$ $\frac{\tan \theta }{\theta }=0$ and $\underset{\theta \to 0}{lim}$ $\frac{\sin \theta }{\theta }=0$

Also $\underset{\theta \to 0}{lim}$ $\frac{{\sin }^n\theta }{{\theta }^n}=0$

Using these we can evaluate the limit to be

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

$=\frac{1}{32}$