Find $\lim_{x \to 0} \frac{\sqrt{1- \cos (2x)}}{\sin(3x)}$

Question

I'm trying to solve this limit:

$$\lim_{x \to 0} \frac{\sqrt{1-\cos(2x)}}{\sin(3x)}.$$

I need to solve this with out using l'Hôpital's rule. How can I solve it? I thought I can use the rule that $$\lim_{x \to 0} \frac{\sin(x)}{x} = 1.$$

This is what I did:

$$ \lim_{x \to 0} \frac{\sqrt{1-\cos(2x)}}{\sin(3x)} = \lim_{x\to 0} \frac{\sqrt{2\left(\frac{1-\cos(2x)}{2}\right)}}{3\left(\frac{\sin(3x)}{3}\right)}\\ =\lim_{x\to 0} \frac{\sqrt{2\cdot 0}}{3\cdot 1}=\lim_{x\to 0} \frac{0}{3} = 0.$$

I know that what I did is wrong, but I can't understand why.

Answer 1

We have

$\frac{\sqrt{1-\cos 2x}}{\sin 3x}=\frac{\sqrt2 |\sin x|}{\sin 3x}\to\frac{\sqrt2}3$ for $x \to 0+0$

and

$\frac{\sqrt{1-\cos 2x}}{\sin 3x}=\frac{\sqrt2 |\sin x|}{\sin 3x}\to-\frac{\sqrt2}3$ for $x \to 0-0$.

Conclusion ?

Answer 2

Hint: use : $\sin kx/ kx \to 1$ as $x \to 0$