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.
No comments:
Post a Comment