Differentiability of $x^\alpha \sin(x^{-\beta})$ at $x = 0$ \begin{align*}
 f(x) = \left\{\begin{array}{ll}   
      0 &  \text{ if } x=0\\
  x^\alpha \sin(x^{-\beta}) &  \text{ otherwise }
 \end{array}\right.
 \end{align*}
Determine the values of $\alpha$ and $\beta$ for which this function is differentiable at $x=0$.
I found the derivative, but I don't know what to do after...
 A: Hint: Remember the definition of the derivative at $0$:
$$f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x} = \lim_{x \to 0} x^{\alpha - 1} \sin\left(x^{-\beta}\right)$$
Now the sine function oscillates as the input goes to zero or to infinity, so the only way the limit can possibly exist is if the term $x^{\alpha - 1}$ kills the oscillation by tending to zero. This suggests that you might want to consider cases based on $\alpha > 1$, $\alpha = 1$ and $\alpha < 1$.
Can you take it from here?
A: Just to clarify some cases. I came too late
$$\lim_{x \to 0}\frac{x^\alpha sin(x^{-\beta})-0}{x-0}=\lim_{x \to 0}x^{\alpha-1} sin(x^{-\beta})$$
Then is differentiable if the previous limit exists.
We have the case $\alpha>1$, $\alpha=1$ $\alpha>1$ and the cases for $\beta=0$,$\beta>0$, $\beta<0$
$\alpha>1$ $\beta=0$ clearly converges to $0$
$\alpha<1$ $\beta=0$ clearly diverges
$\alpha=1$ $\beta=0$ clearly converges
$\alpha>1$ $\beta<0$ converges to $0$
$\alpha<1$ $\beta<0$ we have a limit of the form:
$$\lim_{x \to 0}\frac{sin(x^{-\beta})}{x^{1-\alpha}}$$
Take $u=x^{-\beta}$ then $x^{1-\alpha}=u^{\frac{\alpha-1}{\beta}}$
$$\lim_{u \to 0}\frac{sin(u)}{u^{\frac{\alpha-1}{\beta}}}$$
Here if $\frac{\alpha-1}{\beta}<1$ it converges and in another case diverges
$\alpha=1$ $\beta<0$ clearly converges
