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?

  • 3
    $\begingroup$ $\sqrt{\sin^2(x)}$ does not equal $\sin(x)$. It equals $|\sin(x)|$. $\endgroup$
    – Snaw
    Commented Jan 26, 2022 at 18:47

1 Answer 1


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$$\lim_{x\to0^+}\frac{|\sin x|}x=1\quad\text{whereas}\quad\lim_{x\to0^-}\frac{|\sin x|}x=-1.$$


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