Prove $f$ isn't continuous at $\frac{1}{\pi}$

Question

Let $f(x)=\left\lfloor {\sin {1 \over x}} \right\rfloor$ (meaning floor of $\sin x$).
I need to prove that $f(x)$ isn't continuous at $x=\frac{1}{\pi}$.

Proof:
For a nehiborhood of $\frac{1}{\pi}$:
$x < \frac{1}{\pi} \implies \frac{1}{x} > \pi \implies -1 < \sin \frac{1}{x} < 0 \implies \left\lfloor \sin \frac{1}{x} \right\rfloor = -1$ $x > \frac{1}{\pi} \implies \frac{1}{x} < \pi \implies 0<\sin \frac{1}{x} < 1 \implies \left\lfloor \sin \frac{1}{x} \right\rfloor = 0$

Now, the author claims:

$$\lim\limits_{x\to \frac{1}{\pi}^+} f(x) = 0 \ne \lim\limits_{x\to \frac{1}{\pi}^-} f(x) = 1$$

Didn't he mean:
$$\lim\limits_{x\to \frac{1}{\pi}^+} f(x) = -1 \ne \lim\limits_{x\to \frac{1}{\pi}^-} f(x) = 0$$

So is it a mistake of the author or am I missing something?

Answer 1

Yes, it is a mistake, but only the second limit is wrong (it should be $-1$, as you have calculated).

Recall that $x \to a^+$ means $x \to a$ from the right. So $x \to \dfrac{1}{\pi}^+$ is the case where $x > \dfrac{1}{\pi}$, so the limit is $0$ (as the author has correctly written) and not $-1$ (as you wrote).

The other limit where $x \to \dfrac{1}{\pi}^-$ is where $x < \dfrac{1}{\pi}$, and this is indeed $-1$, not $1$.

So

$$\boxed{\lim\limits_{x \to \frac{1}{\pi}^+} f(x) = 0 \ne -1 = \lim\limits_{x \to \frac{1}{\pi}^-} f(x)}$$