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How would I give a simple example of a function $f$ differentiable in a deleted neighborhood of $x_0$ such that $\lim_{x\to x_0}f^\prime(x)$ does not exist? I cannot seem to think of an example.

A delete neighborhood is an open interval about $x_0$ which does not contain $x_0$. So, $(x_0-\delta,x_0+\delta)-\{x_0\}$ for some $\delta>0$.

How would something be differentiable in a deleted neighborhood if at the point of the derivative, the limit does not exist. Presumably, the derivative ends up looking something like $\lim_{x\to x_0} \dfrac{1}{x}$, if it does not exist.

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  • $\begingroup$ Must be your function continuous? $\endgroup$
    – Dog_69
    Commented Nov 19, 2018 at 13:50
  • $\begingroup$ @Dog_69 No, it can be any function we can dream up $\endgroup$
    – kaisa
    Commented Nov 19, 2018 at 13:51
  • $\begingroup$ I was thinking about the Heaviside's function but I will say the absolute value $|x|$ around $x=0$. $\endgroup$
    – Dog_69
    Commented Nov 19, 2018 at 13:58

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Take $f(x) = x \sin (1/x)$ near $0$

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Classic example: $$\sqrt[3]{(x-x_0)^2}$$

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You may try $f(x)=x^2\cos(1/x)$, so that $f'(x)=2x\cos(1/x)-\sin(1/x)$ has a point of discontinuity at $x=0$.

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    $\begingroup$ This is a slightly better example than mine, in fact, because $f$ is differentiable at zero as well. $\endgroup$
    – Richard Martin
    Commented Nov 19, 2018 at 14:09
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Does $f(x)=x^\frac 12 $ count?
$f'(x)=\frac 1{2x^\frac 12}$ which is discontinuous at $x=0$

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    $\begingroup$ Yes, it does! But see my remark on $x^2 \cos 1/x$ $\endgroup$
    – Richard Martin
    Commented Nov 19, 2018 at 14:10
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$$\ln'(x) = \dfrac{1}{x}$$

If you are looking for that exact derivative.

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