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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!

codetalker
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martina
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  • The limit does not exist –  Nov 15 '16 at 18:15
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    You are right, the one-sided limits exist and are different, so the limit itself doesn't exist. If your textbook says otherwise then find a better textbook. – dxiv Nov 15 '16 at 18:16
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    Your approach is correct. The one-sided limits both exist and are different, so the limit does not exist. – MPW Nov 15 '16 at 18:16
  • Which textbook is that? – Did Nov 15 '16 at 18:18
  • but why https://www.symbolab.com/solver/limit-calculator/%5Clim_%7Bx%5Cto%200%7D%5Cleft(sgnx%5Cright) then ?? – martina Nov 15 '16 at 18:18
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    @martina That limit calculator doesn't understand that "sgn" is a function. It is treating it as the product of three constants, s, g, and n. –  Nov 15 '16 at 18:19
  • @martina This however returns: "No Steps: Steps are currently not supported for this problem". – dxiv Nov 15 '16 at 18:24
  • Thanks. And what is different if I have $\lim\limits_{x\to0} x sgn (x)$? – martina Nov 15 '16 at 18:27
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    @martina I entered it as \sgn (x) and believe that causes symbolab to treat it as the function, but I am not familiar with their syntax. Compare to Wolfram Alpha which does handle the limit correctly here. – dxiv Nov 15 '16 at 18:29
  • On suitable definitions $\operatorname{sgn}0=0$, but $\lim_{x\to0}\operatorname{sgn}x$ still doesn't exist. We say $\operatorname{sgn}$ is discontinuous at $0$. – J.G. Mar 18 '22 at 20:24

2 Answers2

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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.

Michael Hardy
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Q. limx→0sgn(x)

LHS= (0-h)= (-h)= (-1)

RHS= (0+h) = (h) = (1)

LHS=-1 RHS=1 Therefore, LHS is not equal to RHS , limit does not exist .

Parth
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  • As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – Community Mar 18 '22 at 20:15