Let us consider the function
$$f(x)=\begin{cases}x^p\sin\left(\frac{1}{x^q}\right), & \text{ if } x\neq 0\\
0, & \text{ if } x=0
\end{cases}$$
Where $p,q\in\mathbf{R}$.
For which values of $p,q$ the above function is continuous and differentiable.
I have tried in this way:
To show continuity, $\lim\limits_{x\to 0} x^p\sin(\frac{1}{x^q})=\lim\limits_{x\to 0}x^{p-q}x^q \sin(\frac{1}{x^q})$
For differentiability,
$\lim\limits_{x\to 0}\frac{f(x)-f(0)}{x-0}=\lim\limits_{x\to 0} x^{p-1}\sin(\frac{1}{x^q})$
What can I say after this?
