Continuity of trigonometric function with absolute values 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}$?
 A: The case work isn't necessary , because if you see the graph of $y=\sin x$ , you would figure out that whether $x$ be $0^{+}$ or $0^{-}$ , the value of $\sin x $ would give you the value $0$. 
Note that case work is necessary when you're getting different values for the same function from the right hand side and left hand side of the given point about which you've got to check the limit.
A: 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 $0$.
