Question

Solve $\underset{x\to \frac{\pi }{4}}{lim}\frac{f\left(x\right)-f\left(\frac{\pi }{4}\right)}{x-\frac{\pi }{4}}$, where $f\left(x\right)=\sin 2x$

Answer 1

$f\left(\frac{\pi }{4}\right)=\sin \frac{\pi }{2}=1$

$\underset{x\to \frac{\pi }{4}}{lim}\frac{f\left(x\right)-f\left(\frac{\pi }{4}\right)}{x-\frac{\pi }{4}}=\underset{x\to \frac{\pi }{4}}{lim}\frac{\sin 2x-1}{x-\frac{\pi }{4}}$

It is of the form $\frac{0}{0}$

So we will apply L Hospital Rule

$\underset{x\to \frac{\pi }{4}}{lim}\frac{\frac{d}{dx}\left(\sin 2x-1\right)}{\frac{d}{dx}(x-\frac{\pi }{4})}=\frac{2\cos 2x}{1}=2\cos \frac{\pi }{2}=0$