Prove that a function has a derivative at $x=0$ Okay, we have this function:
$f(x)= |x|^\alpha \sin(\frac{1}{x})$, if $x\neq0$
$f(x)= 0$, if $x=0$
The question is, at point x=0:
1) At which value of $\alpha$ does $f(x)$ have a derivative?
2) At which value of $\alpha$ does $f(x)$ have a continuous derivative?
The answer is 
1) $\alpha>1$ 
2) $\alpha>2$
For question 1) I have to prove that this limit exist:
$$\lim_{h\to0} \frac{f(0+h)-f(0)}{h} $$
Which is this:
$$\lim_{h\to0} \frac{|h|^\alpha sin(\frac{1}{h})}{h} $$
We know that $\sin(\frac{1}{h})$ as $h\to0$ doesn't have a limit but it is bounded. So, if we prove that $lim_{h\to0} \frac{|h|^\alpha}{h} = 0 $, then we would have proven that the whole limit is equal to 0, thus, the derivative at point x=0 is equal to 0. Is this reasoning correct?
If $\alpha > 1$, we can say that:
$$\lim_{h\to0} | \frac {h^\alpha}{h} | = \lim_{h\to0} |h^{\alpha-1}| = 0 $$
From this follows that:
$$\lim_{h\to0} \frac{|h|^\alpha}{h}=0 $$
Is this reasoning correct?
For the second question, I am completely lost. This is my first time posting here, and I'd be very thankful if you guys could help me out.

For question $1$ we need that the following limit exists
$$\lim_{h\to0} \frac{|h|^\alpha \sin(\frac{1}{h})}{h}$$
and since $\sin(\frac{1}{h})$ oscillates and is bounded we need that
$$\lim_{h\to0} \frac{|h|^\alpha }{h}=0 \implies \alpha>1$$
For question $2$ we need to check that the derivative is continuous and since we have
$$f'(x)=|x|^{\alpha-2}\left(\alpha x \sin(\frac{1}{x})-\cos(\frac{1}{x})\right)$$
in order to have continuity at the origin we need that as $x \to 0$ we have $f'(x) \to 0$.
