Question

Evaluate:

${lim}_{x\to \frac{\pi }{4}}\frac{\sin x-\cos x}{x-\frac{\pi }{4}}$

Answer 1

${lim}_{x\to \frac{\pi }{4}}\frac{\sin x-\cos x}{x-\frac{\pi }{4}}$ $(\frac{0}{0}\to \text{indeterminate form})$

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

$={lim}_{x\to \frac{\pi }{4}}\frac{\sqrt{2}(\cos \frac{\pi }{4}.\sin x-\sin \frac{\pi }{4}.\cos x)}{x-\frac{\pi }{4}}$

$={lim}_{x\to \frac{\pi }{4}}\frac{\sqrt{2}\sin (x-\frac{\pi }{4})}{(x-\frac{\pi }{4})}$

$(\because \sin (A-B)=\sin A\cos B-\cos A\sin B)$

Put $x-\frac{\pi }{4}=y$

So, when $x\to \frac{\pi }{4}\Rightarrow y\to 0$

$={lim}_{y\to 0}\sqrt{2}\times \frac{\sin y}{y}$

$=\sqrt{2}$ $[\because {lim}_{x\to 0}\frac{\sin x}{x}=1]$