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Let

$f(x) = \begin{cases} |x|^m\sin{\left(\frac{1}{|X|^n}\right)} & x\neq 0\\ 0 & x=0\end{cases}$.

Questions:

  1. For what values of $m,n \in \mathbb{R}$ is $f$ continuous at $0$?
  2. Differentiable at $0$?
  3. Continuously Differentiable at $0$?

My (partly successful) attempt:

  1. $f$ 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.

  2. 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?

  3. 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?

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  • $\begingroup$ For the second question, you can use the first order approximation $\sin(x) = x + o(|x|^2)$ to simplify the problem. Also, you should specify if $n,m$ are natural or relative integers. $\endgroup$ – nicomezi Jan 8 at 8:55
  • $\begingroup$ @nicomezi edited to answer your question. Thanks, but we haven't learned that approximation yet... $\endgroup$ – Mathguy Jan 8 at 9:35
  • $\begingroup$ Thank you for editing. For the last questions, you can just compute the derivative formally, without taking care of the actual value of $m$ and $n$. Once it is done, then you can think further about the values you need for the derivative to be defined (and continuous). $\endgroup$ – nicomezi Jan 8 at 9:39
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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

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

  2. $f$ 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 $0$.

  3. $f$ 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 $m > n+1$.

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