For What $n, m$ is $f(x)$ Differentiable? Continuously Differentiable? Let
$f(x) = \begin{cases} |x|^m\sin{\left(\frac{1}{|X|^n}\right)} & x\neq 0\\ 0 & x=0\end{cases}$.
Questions:


*

*For what values of $m,n \in \mathbb{R}$ is $f$ continuous at $0$?

*Differentiable at $0$?

*Continuously Differentiable at $0$?


My (partly successful) attempt:


*

*$f$ continuous at $0$ precisely when $0<m$ or $n\lt m\leq0$. This is because if $m>0$ then $\lim_{x\to 0}{|x|^m}=0$ and $\sin{\left(\frac{1}{|X|^n}\right)}$ is bounded. Else, if $n<m\leq 0$ then $|x|^{m-n} \frac{\sin{\left(\frac{1}{|X|^{n}} \right)}}{\frac{1}{|X|^n}}$ converges to $0$ as well.

*I know I need to check when the limit $\lim_{x\to 0}{\frac{f(x)}{x}}$ exists. This is too difficult and divided into too many subcases. Is there an easier way to do it which I am missing? How do you divide the cases and show for each case?

*Same as 2. Is there an easier way to do it which I am missing? How do you divide the cases and show for each case?
 A: Note that $f = f_{m,n}$ is an even function, thus it suffices to consider
$$s_{m,n} : [0,\infty) \to \mathbb R, s(x) = \begin{cases} x^m \sin (x^{-n}) & x > 0 \\ 0 & x = 0 \end{cases}$$
This function is continuously differentiable on $(0,\infty)$ with
$$s_{m,n}'(x) = mx^{m-1} \sin (x^{-n}) - nx^{m-n-1}\cos(x^{-n}) = ms_{m-1,n}(x) - nx^{m-n-1}\cos(x^{-n}). $$
Then


*

*$f$ is continuous at $0$ iff $s_{m,n}$ is continuous at $0$ .

*$f$ is differentiable at $0$ iff $s_{m,n}$ is (right) differentiable at $0$ with $s_{m,n}'(0) = 0$ which means that $\lim_{x \to 0+} \dfrac {s_{m,n}(x)}{x} = \lim_{x \to 0+} s_{m-1,n}(x) = 0$, i.e. that $s_{m-1,n}$ is continuous at $0$.

*$f$ is continuously differentiable at $0$ iff $s_{m,n}$ is (right) differentiable at $0$ with $s_{m,n}'(0) = 0$ and $\lim_{x \to 0+} s_{m,n}'(x) = 0$. By 2. this is equivalent to $s_{m-1,n}$ being continuous at $0$ and [ $n = 0$ or $\lim_{x \to 0+} x^{m-n-1}\cos(x^{-n}) = 0$ ].
Case 1: $n = 0$.
Then $s_{m,0}(x) = a x^m$ with $a = \sin 1$. This is continuous at $0$ iff $m > 0$. It is differentiable at $0$ with derivative $0$ iff $m > 1$. It is also continuosly differentiable at $0$ with derivative $0$ iff $m > 1$.
Case 2: $n > 0$.
Then $x^{-n} \to \infty$ as $x \to 0+$ and $\sin(x^{-n})$ oscillates between $-1$ and $1$. Hence $s_{m,n}$ is continuous at $0$ iff $m > 0$ and differentiable at $0$ with derivative $0$ iff $m > 1$. Since also $\cos(x^{-n})$ oscillates between $-1$ and $1$, we see that $s_{m,n}$ is continuously differentiable at $0$ with derivative $0$ iff $m > 1$ and $m - n - 1 > 0$. These two conditions can be summarized to $m > n +1$ (recall $n > 0$).
Case 3: $n < 0$.
Then $x^{-n} \to 0$ as $x \to 0+$ and
$$s_{m,n}(x) = x^{m-n} \dfrac{\sin(x^{-n})}{x^{-n}} .$$
Since $\lim_{x \to 0+} \dfrac {\sin(x^{-n})}{x^{-n}} = 1$ we conclude that
$s_{m,n}$ is  continuous at $0$ iff $m > n$ and differentiable at $0$ with derivative $0$ iff $m > n+1$. It is continuously differentiable at $0$ with derivative $0$ iff $m > n+1$ because this yields $\lim_{x \to 0+} x^{m-n-1}\cos(x^{-n}) = 0$.
Conclusion:
$f$ is continuous iff $m > \min(0,n)$, differentiable at $0$ with derivative $0$ iff $m > \min(0,n) +1$ and continuously differentiable at $0$ with derivative $0$ iff $m > n+1$.
