Discontinuous point for a function $\frac{|\sin{x}|}{\sin{x}}$ I want to determine what type of discontinuity a function has by using one-sided limits for the function $$f(x) = \frac{|\sin{x}|}{\sin{x}}$$
I found the left and right hand limits at $x=0$ (because the $f(x)$ is undefined for $f(0)$). I have found that $$\lim_{x \rightarrow 0^-}f(x) = 0 \qquad \text{and} \qquad \lim_{x \rightarrow 0^+}f(x) = 0$$
It appears to me, that the limit exists and is zero, but it shouldn't be like that I guess. Could someone help me out?

It might prove helpful to visualize the function:

At every integer multiple of $\pi$, there is a discontinuity, since $\sin(k\pi) \equiv 0$ for all integers $k$. We make a jump because the function effectively "flips sign" here: wherever $\sin(x)<0$ we have $f(x) = |\sin(x)|/\sin(x) = -1$ and similarly for $\sin(x) > 0$ we have $f(x) = 1$.
Let's pick $0$ as our discontinuity. We want to show the right- and left-hand limits are not equal there. Indeed, let's consider a path from $x=2$ (or whatever less than $\pi$) to $0$. Since $x>0$ here, then $\sin(x) > 0$ and $|\sin(x)| = \sin(x)$. Then, for a "path" of $x$ going from positive numbers to $0$ we would have
$$f(x) = \frac{|\sin(x)|}{\sin(x)} = \frac{\sin(x)}{\sin(x)} = 1 \xrightarrow{x \to 0^+} 1$$
However, let's say we started at some $x$ less than $0$, say $x=-2$, and approached $0$ for the left-hand limit. Then $\sin(x) < 0$ on this interval, and $|\sin(x)| =- \sin(x)$ as a result. And thus here,
$$f(x) = \frac{|\sin(x)|}{\sin(x)} = \frac{-\sin(x)}{\sin(x)} = -1\xrightarrow{x \to 0^-} -1$$
We in turn conclude:
$$\lim_{x \to 0^-} f(x) = -1 \;\;\;\;\; \lim_{x \to 0^+} f(x) = 1$$
establishing a discontinuity at $x=0$. A similar argument could be done for any $x=k\pi$.
