Question

Evaluate the limit:
$\underset{x\to \pi }{lim}\frac{1-\sin \frac{x}{2}}{\cos \frac{x}{2}(\cos \frac{x}{4}-\sin \frac{x}{4})}$

Answer 1

We have:
$\underset{x\to \pi }{lim}\frac{1-\sin \frac{x}{2}}{\cos \frac{x}{2}(\cos \frac{x}{4}-\sin \frac{x}{4})}$

Rationalise the Numerator, we get

$=\underset{x\to \pi }{lim}\frac{1-\sin \frac{x}{2}}{\cos \frac{x}{2}(\cos \frac{x}{4}-\sin \frac{x}{4})}\times \frac{1+\sin \frac{x}{2}}{1+\sin \frac{x}{2}}\times \frac{\cos \frac{x}{4}+\sin \frac{x}{4}}{\cos \frac{x}{4}+\sin \frac{x}{4}}$

$=\underset{x\to \pi }{lim}\frac{(1-\sin \frac{x}{2})(1+\sin \frac{x}{2})}{\cos \frac{x}{2}(\cos \frac{x}{4}-\sin \frac{x}{4})(\cos \frac{x}{4}+\sin \frac{x}{4})}\times \frac{\cos \frac{x}{4}+\sin \frac{x}{4}}{1+\sin \frac{x}{2}}$

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

$=\underset{x\to \pi }{lim}\frac{{\cos }^2\frac{x}{2}}{\cos \frac{x}{2}\left(\cos \frac{x}{2}\right)}$ $\times \frac{\cos \frac{x}{4}+\sin \frac{x}{4}}{1+\sin \frac{x}{2}}$
$[\because {\cos }^{\theta }-{\sin }^2\theta =\cos 2\theta ]$

$=\underset{x\to \pi }{lim}\frac{{\cos }^2\frac{x}{2}}{{\cos }^2\frac{x}{2}}$ $\times \frac{\cos \frac{x}{4}+\sin \frac{x}{4}}{1+\sin \frac{x}{2}}$

$=\underset{x\to \pi }{lim}\frac{\cos \frac{x}{4}+\sin \frac{x}{4}}{1+\sin \frac{x}{2}}$

$=\frac{\cos \frac{\pi }{4}+\sin \frac{\pi }{4}}{1+\sin \frac{\pi }{2}}$

$=\frac{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}{1+1}$

$=\frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}}$

Therefore,
$\underset{x\to \pi }{lim}\frac{1-\sin \frac{x}{2}}{\cos \frac{x}{2}(\cos \frac{x}{4}-\sin \frac{x}{4})}=\frac{1}{\sqrt{2}}$