Question

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

Answer 1

Consider the given expression,

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

Expressing tan in the form of sin and cos, we get

${lim}_{x\to \frac{\pi }{2}}\frac{(\cos \frac{x}{2}-\sin \frac{x}{2})(1-\sin x)}{(\cos \frac{x}{2}+\sin \frac{x}{2})(\pi -2x^3)}$

Multiplying and dividing by $(\cos \frac{x}{2}+\sin \frac{x}{2})$ , we get

$={lim}_{x\to \frac{\pi }{2}}\frac{{(\cos \frac{x}{2}-\sin \frac{x}{2})}^2(1-\sin x)}{({\cos }^2\frac{x}{2}+{\sin }^2\frac{x}{2})(\pi -2x^3)}$

$={lim}_{x\to \frac{\pi }{2}}\frac{({\cos }^2\frac{x}{2}+{\sin }^2\frac{x}{2}-2\sin \frac{x}{2}\cos \frac{x}{2})(1-\sin x)}{({\cos }^2\frac{x}{2}+{\sin }^2\frac{x}{2})(\pi -2x^3)}$

$={lim}_{x\to \frac{\pi }{2}}\frac{(1-\sin x)(1-\sin x)}{(1-{\sin }^{2}x)(\pi -2x^3)}$

$={lim}_{x\to \frac{\pi }{2}}\frac{(1-\sin x)}{\cos x(\pi -2x^3)}$

let $\pi -2x=t$

when $x\to \frac{\pi }{2},t\to 0$

$={lim}_{t\to 0}\frac{{(1-\sin (\frac{\pi }{2}-\frac{t}{2}))}^2}{\cos (\frac{\pi }{2}-\frac{t}{2})(\pi -2{(\frac{\pi }{2}-\frac{t}{2})}^3)}$

$={lim}_{t\to 0}\frac{{(1-\cos \frac{t}{2})}^2}{\sin \frac{t}{2}\left(t^3\right)}$

$={lim}_{t\to 0}\frac{4{\sin }^3\frac{t}{4}}{\cos \frac{t}{4}\left(t^3\right)}$

$={lim}_{t\to 0}\frac{4{\sin }^3\frac{t}{4}}{{\left(\frac{t}{4}\right)}^3\times 64}$

$=\frac{4}{64}=\frac{1}{16}$