I have to evaluate the continuity of this function $$ f(x) = \begin{cases} \sin x + \dfrac{\sqrt{1-\cos2x}}{\sin x}, & x \neq 0 \\[6px] \sqrt{2}, & x=0 \end{cases} $$

For the function to be continuous we need $$ \lim_{x \to 0}f(x) = f(0) $$

We know that $ 1 - \cos2x = 2\sin^2x $ so we can rewrite the $ x \neq 0 $ part as: $$ \sin x + \frac{\sqrt{2\sin^2x}}{\sin x} = \sin x + \sqrt{2}\frac{\lvert\sin x\rvert}{\sin x}$$

Do I have to take cases here ($x \to 0^-$ and $x \to 0^+) $ for the $\lvert\sin x\rvert$ to find the limit I want, or can I assume that $ \sin x > 0 $?

**UPDATE**

Taking cases we have: $$\lvert\sin x\rvert = \begin{cases}-\sin x, & x < 0 \\[4px] \sin x, & x>0 \end{cases} $$

so $$ \lim_{x \to 0^-}\sin x + \sqrt{2}\frac{-\sin x}{\sin x} = -\sqrt{2} $$

$$ \lim_{x \to 0^+}\sin x + \sqrt{2}\frac{\sin x}{\sin x} = \sqrt{2} $$

Can I assume something like this?

This Wiki link makes it clear that we can have $-{\sin x}$. Does it have anything to do with the fact that $ n \in \mathbb{Z}$?