Question

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

Answer 1

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

Now, let $x-\pi =t$ , At $x\to \pi$, $t\to 0$

$\Rightarrow x=\pi +t$

So, ${lim}_{t\to 0}\frac{1-sin\frac{\pi +t}{2}}{cos\frac{\pi +t}{2}(cos\left(\frac{\pi +t}{4}\right)-sin\left(\frac{\pi +t}{4}\right))}$

$={lim}_{t\to 0}\frac{1-cos\left(\frac{t}{2}\right)}{-sin\left(\frac{t}{2}\right)(cos\left(\frac{\pi +t}{4}\right)-sin\left(\frac{\pi +t}{4}\right))}$

Simplifying,

$cos\left(\frac{\pi +t}{4}\right)-sin\left(\frac{\pi +t}{4}\right)=cos\left(\frac{\pi }{4}\right)cos\left(\frac{t}{4}\right)-sin\left(\frac{\pi }{4}\right)sin\left(\frac{t}{4}\right)-sin\left(\frac{\pi }{4}\right)cos\left(\frac{t}{4}\right)-cos\left(\frac{\pi }{4}\right)sin\left(\frac{t}{4}\right)=-sin\left(\frac{t}{4}\right)(sin\frac{\pi }{4}+cos\frac{\pi }{4})=-\sqrt{2}sin\left(\frac{t}{4}\right)$

Now, ${lim}_{t\to 0}\frac{1-cos\left(\frac{t}{2}\right)}{\sqrt{2}sin\left(\frac{t}{2}\right)sin\left(\frac{t}{4}\right)}={lim}_{t\to 0}\frac{2{\sin }^2\left(\frac{t}{4}\right)}{\sqrt{2}sin\left(\frac{t}{2}\right)sin\left(\frac{t}{4}\right)}(1-cos2A=2{\sin }^{2}A)={lim}_{t\to 0}\frac{\sqrt{2}\frac{{\sin }^2\left(\frac{t}{4}\right)}{{\left(\frac{t}{4}\right)}^2}\times {\left(\frac{t}{4}\right)}^2}{\left(\frac{sin\left(\frac{t}{2}\right)}{\left(\frac{t}{2}\right)}\right)\times \left(\frac{t}{2}\right)\times \left(\frac{sin\left(\frac{t}{4}\right)}{\left(\frac{t}{4}\right)}\right)\times \left(\frac{t}{4}\right)}=\frac{\sqrt{2}\times 8}{16}\left({lim}_{x\to 0}\frac{sinx}{x}=1\right)=\frac{1}{\sqrt{2}}$