Question

The value of $x\stackrel{lim}{\to }\infty x\sin \frac{\pi }{4x}\cos \frac{\pi }{4x}$

Answer 1

${lim}_{x\to \infty }x\sin \left(\frac{\pi }{4x}\right)\cos \left(\frac{\pi }{4x}\right){lim}_{x\to \infty }\frac{\sin \left(\frac{\pi }{4x}\right)\cos \left(\frac{\pi }{4x}\right)}{\frac{1}{x}}$

Let $\frac{1}{x}=t$ then as $x\to \infty ,t\to 0$

$\therefore {lim}_{t\to 0}\frac{2\sin \left(\frac{\pi t}{4}\right).\cos \left(\frac{\pi t}{4}\right)}{2t}{lim}_{t\to 0}\frac{\sin \frac{\pi t}{2}}{2t}$

We know that

${lim}_{x\to 0}\frac{\sin x}{x}=1$

Using this ,

${lim}_{t\to 0}\frac{\sin \left(\frac{\pi }{2}\right)t}{2t}={lim}_{t\to 0}\frac{\sin \left(\frac{\pi }{2}\right)t}{2\left(\frac{\pi }{2}t\right).\left(\frac{2}{\pi }\right)}=\frac{\pi }{4}{lim}_{t\to 0}\frac{\sin \left(\frac{\pi }{2}t\right)}{\frac{\pi t}{2}}=\frac{\pi }{4}$