Question

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

Answer 1

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

Let $t=x-\frac{\pi }{2}$

$\underset{t\to 0}{lim}\frac{1-\cos t}{-8t^3}\tan (-\frac{t}{2})=\frac{1}{8}\underset{t\to 0}{lim}\frac{1-\cos t}{t^2}\times \frac{\tan \frac{t}{2}}{t}=\frac{1}{8}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{32}$