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I have the function

$$ f: [0, 1] \rightarrow \mathbb{R}, \quad x \mapsto \begin{cases} x^\alpha \sin \frac{1}{x} && x > 0 \\ 0 && x=0 \end{cases} $$

and I need to proof that $f'$ is not continous on $[0, 1]$ for $\alpha \in (1, 2]$. I've already shown that it is continous for $\alpha > 2$ and I know that it is differentiable on $[0, 1]$ if and only if $\alpha > 1$.

I know that if goes wrong for $f'(0)$.

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If $f'$ is continous on $[0,1]$ it's especially continous in 0 but $$f'(x) = ax^{a-1}\sin\frac{1}{x} - x^{a-2}\cos\frac{1}{x}$$ for $x>0$ and because $$\lim_{x \searrow 0}\; f'(x)$$ doesn't exists for $a \in (1,2]$ the derivative $f'$ is not continuous in $0$ hence in $[0,1]$.

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    How do you know that $\lim_{x \searrow 0}\; f'(x)$ does not exist? How to **proof** that? – 2017-01-31
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    The limit of the first summand exists and is $0$ due to $$\left|\sin\frac{1}{x}\right| \le 1$$. So if the whole limit should exist also the second part should have a limit but it doesnt. To proof it compare the limit for $x_n = (2\pi n)^{-1}$ which is infinite and $x_n = \left(2\pi n + \frac{\pi}{2}\right)^{-1}$ which is $0$. So the whole limit doesn't exist. – 2017-02-01