Prove: For which $\alpha\geq0$ is the following differentiable. $f: \mathbb R \to \mathbb R$
$f(x): \begin{cases} |x|^{\alpha}sin(\frac{1}{x}) & x\neq 0 \\ 0 & x = 0 \end{cases}$
I am fine with $\alpha \in \mathbb R_{\geq1}\cup0$ , but I am struggling on $\alpha \in (0,1)$
Let's take $\sqrt{x}sin(\frac{1}{x})$ for example.
It follows $\frac{f(x)-f(0)}{x-0}=\frac{\sqrt{x}sin(\frac{1}{x})}{x}=\frac{sin(\frac{1}{x})}{\sqrt{x}}$ I tried using l'hopital and it does not work.
When $\alpha \in (0,1)$, look at what happens as $x$ gets close to $0$.
In particular (for any $\alpha$), look at the sequence $x_n = \frac{2}{\pi(4n+1)}$ for $n\in\mathbb{N}$. Then, $D_f(x_n) = \frac{f(x_n)-f(0)}{x_n-0} = \left|\frac{(4n+1)\pi}{2}\right|^{1-\alpha} \sin(\frac{(4n+1) \pi}{2}) = \left|\frac{(4n+1)\pi}{2}\right|^{1-\alpha}$. If $D_f(x_n)$ diverges as $n\rightarrow\infty$, then $f$ is not differentiable at $x=0$.
Note: The converse is not necessarily true $-$ if $D_f(x_n)$ converges, it doesn't necessarily mean that $f$ is differentiable at $x=0$.