# How to give an example of a $f$ differentiable in a deleted neighborhood of $x_0$ such that $\lim_{x\to x_0}f^\prime(x)$ doesn't exist?

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.

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

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

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

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 $$x=0$$.

• This is a slightly better example than mine, in fact, because $f$ is differentiable at zero as well. Nov 19, 2018 at 14:09

Does $$f(x)=x^\frac 12$$ count?
$$f'(x)=\frac 1{2x^\frac 12}$$ which is discontinuous at $$x=0$$

• Yes, it does! But see my remark on $x^2 \cos 1/x$ Nov 19, 2018 at 14:10

$$\ln'(x) = \dfrac{1}{x}$$

If you are looking for that exact derivative.