Question

Find:

$\underset{x\to 0}{lim}\frac{\sqrt{1-\cos 2x}}{\sqrt{2}x}$

Answer 1

$\begin{array}{c}\underset{x\to 0}{lim}\frac{\sqrt{1-\cos 2x}}{\sqrt{2}x}\\ \underset{x\to 0}{lim}\frac{\sqrt{2}\left|\sin x\right|}{\sqrt{2}}\\ \underset{x\to 0}{lim}\frac{\left|\sin x\right|}{x}\end{array}$

RHS

$\begin{array}{c}x=0+h\\ \underset{h\to 0}{lim}\frac{\sinh }{h}=1\end{array}$

LHS

$\begin{array}{c}x=0-h\\ \underset{h\to 0}{lim}-\frac{\sinh }{-h}=1\end{array}$

So,

$\underset{x\to 0}{lim}\frac{\left|\sin x\right|}{x}=1$