I've been asked to find the limit, without using L'hospital's rule, of:
$$\lim\limits_{x\to0}\frac{\sqrt{1-\cos(2x)}}{x}$$
Here's my attempt:
$$\begin{aligned}&\lim\limits_{x\to0}\frac{\sqrt{1-\cos(2x)}}{x}\\=&\lim_{x\to0}\frac{\sqrt{2\sin^2(x)}}{x}\\=&\lim_{x\to0}\frac{\sqrt{2}\sin(x)}{x}\\=&\sqrt{2}\end{aligned}$$
So my question is what's the problem here? The graph shows that limit doesn't exist. In which situations do we have to find LHL and RHL?
Best Answer
The problem lies in the equality $\sqrt{\sin^2(x)}=\sin(x)$, which is false. What you have is $\sqrt{\sin^2(x)}=\bigl|\sin(x)\bigr|$ instead. And you have$$\lim_{x\to0^+}\frac{|\sin x|}x=1\quad\text{whereas}\quad\lim_{x\to0^-}\frac{|\sin x|}x=-1.$$