We define the function $f$ in the following manner:
$$f(x) = \begin{cases} 0, & \text{if $x = 0$} \\ \vert x \vert^\alpha \text{sin}(\frac{1}{x}), & \text{if $x \neq 0$} \end{cases}$$
Prove that the function $f$ is differentiable at $x=0$ if and only if $\alpha > 1$
($\Rightarrow$) Suppose the function $f$ is differentiable at $x=0$, then the limit: $$\lim_{x \rightarrow 0 } \frac{f(x)-f(0)}{x-0}=\lim_{x \rightarrow 0 } \frac{\vert x\vert^\alpha \text{sin}(\frac{1}{x})}{x}$$
must exist, I claim that the limit exists and is equal to $0$ when $\alpha > 1$
Case (1): $x>0$, we have: $$\lim_{x \rightarrow 0 } x^{\alpha-1} \text{sin}(\frac{1}{x})$$
Idea: Prove that $\lim_{x \rightarrow 0 } x^{\alpha-1} =0$ if and only if $\alpha > 1$, then apply Squeeze Theorem.
Hence for every $\epsilon>0$, choose $\delta=\epsilon ^ {\frac{1}{\alpha - 1}}$(This can be done because $\alpha > 1$), then: $$0<\vert x\vert<\delta \Rightarrow\vert x\vert^{\alpha -1}<\delta^{\alpha - 1}=\epsilon$$
Hence by Squeeze Theorem, I must have the limit being equal to $0$.
Case (2) is similar except that it is for $x<0$
Not sure if I missed out anything though, appreciate it if you can provide me with some feedback on this proof!
Any idea on how to do $(\Leftarrow)$ is greatly appreciated, or is it actually trivial? (As in I can deduce it from ($\Rightarrow$))
To show $\leftarrow$ note that $x^{\alpha-1}$ diverges as $x\to 0^+$ and $\alpha<1$. Also $\sin\dfrac{1}{x}$ has no limit when $x\to 0$ therefore the limit $x^{\alpha -1}\sin\dfrac{1}{x}$ doesn't exist if $\alpha \le 1$. So it should be that $\alpha >1$. If so we can write $$\lim_{x\to 0^+}x^{\alpha-1}\sin \dfrac{1}{x}=\lim_{x\to \infty}\dfrac{\sin x}{x^{\alpha-1}}=0$$this means that your approach is correct. The same is true for $x<0$ since the function is odd.