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I am solving the following problem:

I am to consider a function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x)=x^2sin(\frac{1}{x})$ when $x \neq0$ and $f(x)=0$ when $x=0$.

And I am to show that the derivative f'(x) exists for any x $\in \mathbb{R}$, but $f' : \mathbb{R} \to \mathbb{R}$ is not a continuous function.

My strategy is to compute the derivative at any point and inspect if the formula is continuous or not.

So for the derivative I get f'(x)=2xsin$(\frac{1}{x^{2}}) - cos(\frac{1}{x^{2}})$.

Thereafter id like to inspect what happens with f'(x) when f'(x)=0 by using $\epsilon-\delta$ in some way to find a there is no limit so that I have a function that is not continuous.

My problem is that I don't know how to use the $\epsilon-\delta$ definition to prove that the function is not continuous. Any help would be appreciated, thanks.

Lavorizia Vaughn
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    The limit $\displaystyle \lim_{x \to 0} \Big( 2x\sin\big(\tfrac{1}{x^2}\big)-\cos\big(\tfrac{1}{x^2}\big) \Big)$ doesn't exists, so, no matter what $f'(0)$ is, $f'$ cannot be continuous at $0$. – azif00 Apr 16 '21 at 23:27
  • That makes total sense, but what bothers me is the fact that this part of the proof can be worked out using the $\epsilon-\delta$ definition. I would really appreciate some help showing that. – Lavorizia Vaughn Apr 17 '21 at 20:47

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