Question

Evaluate the limit:
$\underset{x\to \frac{\pi }{4}}{\text{lim}}\frac{f\left(x\right)-f\left(\frac{\pi }{4}\right)}{x-\frac{\pi }{4}}$, where $f\left(x\right)=\sin 2x$

Answer 1

We have,

$\underset{x\to \frac{\pi }{4}}{\text{lim}}\frac{f\left(x\right)-f\left(\frac{\pi }{4}\right)}{x-\frac{\pi }{4}}$

$=\underset{x\to \frac{\pi }{4}}{\text{lim}}\frac{\sin 2x-\sin \frac{\pi }{2}}{x-\frac{\pi }{4}}$

$=\underset{x\to \frac{\pi }{4}}{\text{lim}}\frac{\sin 2x-1}{x-\frac{\pi }{4}}$

Put $x=\frac{\pi }{4}+h,h\to 0$

$=\underset{h\to 0}{\text{lim}}\frac{\sin 2(\frac{\pi }{4}+h)-1}{(\frac{\pi }{4}+h)-\frac{\pi }{4}}$

$=\underset{h\to 0}{\text{lim}}\frac{\sin (\frac{\pi }{2}+2h)-1}{h}$

$[\because \sin (\frac{\pi }{2}+\theta )=\cos \theta ]$

$=\underset{h\to 0}{\text{lim}}\frac{\cos 2h-1}{h}$

$=\underset{h\to 0}{\text{lim}}\frac{-(1-\cos 2h)}{h}$

$=\underset{h\to 0}{\text{lim}}\frac{-2{\sin }^{2}h}{h}$ $[\because 1-\cos 2\theta =2{\sin }^2\theta ]$

$=-2\underset{h\to 0}{\text{lim}}\frac{\sin h}{h}\times \sin h$ $[\because \underset{h\to 0}{\text{lim}}\frac{\sin h}{h}=1]$

$=-2\times 1\times \sin 0$

$=0$

Therefore,

$\underset{x\to \frac{\pi }{4}}{\text{lim}}\frac{f\left(x\right)-f\left(\frac{\pi }{4}\right)}{x-\frac{\pi }{4}}=0$