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}$?
You can't assume $\sin x>0$; what's true is that, in a punctured neighborhood of $0$ one has $\lvert\sin x\rvert>0$, which is quite different.
Just observe that $\lvert-1\rvert>0$, but surely $-1$ is not greater than $0$.
Computing the limits from the left and from the right isn't necessary: what you need to see is whether $\lim_{x\to0}f(x)=\sqrt{2}$. How to do this may involve computing the two limits.
You wouldn't need case work for $$ g(x)=\begin{cases} 1 & x\ne0\\[4px] -1 & x=0\end{cases} $$ would you?
In your particular case it is convenient to do it, because, as you show, $$ \lim_{x\to0^-}f(x) \neq \lim_{x\to0^+}f(x) $$ so the limit at $0$ doesn't even exist and therefore the function is not continuous at $0$.