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Let $k \in \mathbb{N}$ and let $f_{k}$ be a function $f_{k}: \mathbb{R} \rightarrow \mathbb{R}$ with $f_{k}(x)= \begin{cases} x\neq 0 \ f_{k}(x)=\ x^{k} \sin(\frac{1}{x}) \\ x=0 \ \ f_{k}(x)=0 \end{cases}$ For which $k$ is the function differentiable?

Hey guys I'm preparing for exams and I'm trying to improve my proof-writing. Is this one correct?

By definition the derivative is characterized as $\lim_{h \rightarrow 0} \frac{f_{k}(h)-f(0)}{h}$. Thus the the question is equivalent to: For which $k$ does the limit $\lim_{h \rightarrow 0} h^{k-1} \sin(\frac{1}{h})$ exist?

Let $k>1$. Notice that the exponent in the expression $h^{k-1}$ is positive due to our assumption. Further notice that $-1 \leq \sin(\frac{1}{h}) \leq 1$. We multiply the inequality with $h^{k-1}$ and find $h^{k-1} \leq h^{k-1} \sin(\frac{1}{h}) \leq h^{k-1}$. Since $\lim_{h \rightarrow 0} -h^{k-1}=0$ and $\lim_{h \rightarrow 0} h^{k-1}=0$ we know by using the squeeze theorem that $\lim_{h \rightarrow 0} h^{k-1} \sin(\frac{1}{h})=0$. Not only have we shown the existance, we even calculated the limit explicitly. Thus for $k>1$ the function is differentiable.

Let $k=1$ and one can check by using this that the limit becomes $\lim_{h \rightarrow 0} f_k(h)=\lim_{h \rightarrow 0} \sin(\frac{1}{h})$. Assume the limit existed and take the two sequences $x_n:=\frac{1}{2 \pi n+ \pi/2}$ and define $y_n:=\frac{1}{2 \pi n}$ with $x_n, y_n \rightarrow 0$ for $n \rightarrow \infty$. then we find $\sin(\frac{1}{x_n})=\sin(2 \pi n+ \pi/2)=1$ but also $\sin(\frac{1}{y_n})=\sin(2 \pi n)=0$. But if the limit existed the values ​​should match, but they don't. So for $k=1$ not differentiable.

Let $k < 1$. Now notice that we can rewrite the limit $\lim_{h \rightarrow 0} \frac{\sin(\frac{1}{h})}{h^{1-k}}$. Our assumption assures that the exponent $1-k$ is always postitive. Assume the limit existed . Thus we find by substituting $x_n$ that $\frac{\sin(\frac{1}{x_n})}{x_n^{1-k}}=\sin(2 \pi n + \pi/2)(2 \pi n + \pi /2)^{1-k}$. But $\sin(2\pi n + \pi/2)=1$ and thus we find that the limit with the sequence doesn't even exist. So for $k < 1$ not differentiable.

So the function $f_k$ is differentiable if and only if $k>1$

Gary
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Iwan5050
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1 Answers1

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You don't need the case $k<1$, because you assume $k\in\mathbb{N}$. Actually, if $k$ is not an integer then $x^k$ might not even be defined for a negative $x$.

Your proof for the case $k=1$ is correct. As for $k\geq 2$ I have one comment. You wrote that $-1\leq \sin(\frac{1}{h})\leq 1$ implies $-h^{k-1}\leq h^{k-1}\sin(\frac{1}{h})\leq h^{k-1}$. This is true only if $h>0$, and when $h<0$ you have to reverse the inequality. Everyone is aware of it, yet lots of people make this mistake. (here it's very easy to fix, but sometimes it leads to much more serious problems). In order to avoid it, many times it's good to work with absolute value. Remember that $\lim_{x\to x_0}f(x)=0$ if and only if $\lim_{x\to x_0} |f(x)|=0$. So here we have:

$0\leq |h^{k-1}|\cdot|\sin(\frac{1}{h})|\leq |h^{k-1}|\to 0$

And so by the squeeze theorem indeed $h^{k-1}\sin(\frac{1}{h})$ tends to $0$.

Mark
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  • Ok this critic makes total sense. I think I'm blind xD. Thank you very much – Iwan5050 Feb 01 '22 at 08:32
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    @Iwan5050 I'm teaching Calculus $1$ for a few years now, so I know these common mistakes very well. I lost count how many times I told my students about this one. – Mark Feb 01 '22 at 08:34
  • .. for $k \geqslant 2$ and $k$ even will be reverse inequality .. – zkutch Feb 01 '22 at 08:55
  • @zkutch Yes, you are right. But I think simply using absolute value is better than splitting into even and odd cases. – Mark Feb 01 '22 at 08:58