limits - Does $limlimits_{xto0}operatorname{sgn} (x)$ exist?

I have a problem with this exercise

Does this limit exist?

$$\lim\limits_{x\to0} \operatorname{sgn} (x)$$

this limit should exist and its value is $0$ according to our textbook. It is also written, that we can prove it by using one-sided limits.
And there is a problem, because as I see it

$$\lim_{x\to0^-} \operatorname{sgn} (x) = -1$$

$$\lim_{x\to0^+} \operatorname{sgn} (x) = 1$$

(Because the limit goes very close to $0$, but it never reaches it. I also think it is very similar to prove of non-existence $\displaystyle \lim_{x\to0} \sin\frac 1 x$)

I also tried online limit calculators and they said, that one-sided limits equals $0$.

Could you help me find a problem in my approach?

Thanks for your time!

Answer

If the book says the limit is $0$, then it is wrong.

If $\lim\limits_{x\to0+}$ and $\lim\limits_{x\to0-}$ both exist (as finite numbers) and are not equal to each other, then $\lim\limits_{x\to0}$ does not exist.

In some contexts, it might make sense to say it exists as a "principal value", taking an average: $\displaystyle \frac 1 2 \left( \lim_{x\to0+} + \lim_{x\to0-} \right),$ but that is not what is conventionally done when the concept of limit is first introduced, and I would allow is only when the context for it has been explicitly set.