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e.g. a function, f(x) where f(0) approaches 0 from positive x and 0 from negative x, but when you actually compute f(0) it is defined as a different value like 1? And I don't mean some function where you define it differently for x > 0, x < 0 and x = 0.

Additionally a function where the limit at an x approaches a different number from positive x than from negative x, but is actually defined at that point. When the limit approaches a different value from two different directions usually we say it is therefore undefined at that point, but is this always the case?

icantcode
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    What about $f(x) = \frac{x}{x}$? (Assuming "actually compute f(x) it is undefined" was intended to mean "actually compute f(0) it is undefined".) – Dave L. Renfro Aug 08 '21 at 16:21
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    How about $f:\mathbb R\setminus {0}\to \mathbb R, x\mapsto x$? I think the question is unclear you can't possibly mean something like that? – Lukas Betz Aug 08 '21 at 16:21
  • Also note that functions always have a domain. If $0$ is in the domain, then you must have a value $f(0)$. If $0$ is not in the domain then it doesn't make sense trying to "compute" $f(0)$. Some of the answers seem to forget that. – Lukas Betz Aug 08 '21 at 16:52
  • $f(x) = \frac{x^2}{x}$ or $f(x) = \frac{x^4}{x^2}$, or any variations (e.g. see nimmy's answer). – user2661923 Aug 08 '21 at 16:58
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    @DaveL.Renfro Your $f(x)$ does not approach zero when $x$ tends to zero. – Aman Kushwaha Aug 08 '21 at 16:59
  • @Aman Kushwaha: Oops! I missed that part of the requirement. The awkwardness of how the question was asked had me thinking (when I quickly read it, then wrote my comment) that $x$ was approaching $0$ from the left and approaching $0$ from the right, with each of the limits being equal (but not otherwise specified). – Dave L. Renfro Aug 08 '21 at 17:26
  • @icantcode maybe also look here: https://math.stackexchange.com/questions/1525054/why-are-removable-discontinuities-even-discontinuities-at-all – justabit Aug 08 '21 at 17:50
  • I realised I actually wrote this question wrong. Due to the domain issues you are all talking about this question does not make sense. What I meant was that when you compute f(x) at that value it is some other value than the limit as a-->x. So it is still defined at that value but it is just different than what you would have expected looking at the limits. I will edit the question. – icantcode Aug 09 '21 at 15:46
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    I guess that the OP wants a function which has a removable singularity at $0$ but, in some sense, occurs naturally rather than is artificially constructed using a piecewise definition. – badjohn Aug 09 '21 at 16:29
  • @badjohn Yes. – – icantcode Aug 09 '21 at 19:37
  • The Dirac delta function is a near miss. https://en.m.wikipedia.org/wiki/Dirac_delta_function – badjohn Aug 09 '21 at 21:20

3 Answers3

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Let $f:\mathbb R\setminus \{0\}\to \mathbb R$ be a function such that $f(x)=0$.

This function meets all your requirements. First, $f(x)\rightarrow 0$ when $x \rightarrow 0^-$ and $f(x)\rightarrow 0$ when $x \rightarrow 0^+$ . Second, when you compute $f(x)$ at $x=0$ (which you are not supposed to do) i.e., $f(0)$, it is undefined because I defined the function in such a domain that it is undefined at $x=0$. Third, it is not defined differently for $x>0$ and $x<0$.

Is it defined differently for $x=0$? Absurd question! You yourself said $f$ should not be defined at $x=0$ but it takes values close to zero when $x$ approaches zero. So I need to define the function in $\{0-\epsilon, 0+\epsilon\} \setminus \{0\}$, where $\epsilon >0$ is any arbitrary real number

Aman Kushwaha
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What do you mean by "is not defined"? We can always define a function on a subset. If you mean as in forbidden because dividing by 0, then the $f(x) = x+4 = (x^2+4x)/x$ is a good example for a discontinuity. But maybe what you are looking for is $f(x) = \sin(1/|x|)$ , which does not behave well at 0.

justabit
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  • What is the domain of your first function? I can't think of any where this would have a discontinuity. – Lukas Betz Aug 08 '21 at 16:54
  • The domain of $f(x) = x^2+4x)/4$ is $\mathbb{R} \backslash { 0 }$ as you can not divide by zero. But it is a removable discontinuity as everywhere where f is defined it is $x+4$ one could just define it as $f(0) := 4$ . – justabit Aug 08 '21 at 16:59
  • First: the domain is only $\mathbb R\setminus {0}$ if you define it that way it is not some kind of intrinsic property of the expression you gave which by itself is not a function. If you take this domain then this function doesn't have a discontinuity. – Lukas Betz Aug 08 '21 at 17:13
  • Yes, that is the case, as also stated in the answer with $f(x) = 0$. But I think the question of icantcode revolves more around closed forms rather than the definition of domains, but maybe I'm wrong. – justabit Aug 08 '21 at 17:27
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I just thought of this one: $y=0^{|x|}$. From this you can then have $y=0^{|x|} + x^2$ and so on.

icantcode
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