# 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?

• You can't get limit of $0+$ or $0-$, limit should be a number. How did you get this value of limit? Commented May 27, 2019 at 21:58

Actually, since $$\sin x>0$$ when $$x$$ is small and positive and $$\sin x<0$$ when $$x$$ is small and negative,$$\lim_{x\to0^+}\frac{\lvert\sin x\rvert}{\sin x}=1\text{ and }\lim_{x\to0^-}\frac{\lvert\sin x\rvert}{\sin x}=-1.$$However, that's not relevant for the continuity of your function, since $$0$$ does not belong to its domain (which is $$\mathbb R\setminus\pi\mathbb Z$$). And your function is continuous at every point if its domain.

$$0 $$-\theta

Since the left limit is $$-1$$ and right limit is $$1$$, limit on point $$0$$ does not exist.

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 $$x=k\pi$$.