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I was given this function:

$$f(x)=\begin{cases}\displaystyle|x|^p\cos\Big(\frac\pi{|x|^q}\Big),&x\ne0\\0,&x=0\end{cases}$$

And was asked to find for what $p, q>0$ it is differentiable at $x=0$.

First I saw it is continuous when $p>0, q>0$.

Now, I tried to see if the limit for $f'(x)$ exists at $x=0$. This function is even so I looked at the right side only.

$$\lim_{x\to0^+}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0^+}\frac{x^p\cos\Big(\displaystyle\frac\pi{x^q}\Big)}x=\lim_{x\to0^+}x^{p-1}\cos\Big(\frac\pi{x^q}\Big)$$

I get that this limit exists when $p>1$, and for all $q>0$, but looking at the graph online it doesn't seem to be right. What am I doing wrong here (if anything)?

Thanks a lot!

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  • $\begingroup$ You have solved it correctly. Graphical aids are often inaccurate for plots like these. How were you able to draw any conclusion for $f$ near $0$ using its graph? $\endgroup$ – Shubham Johri Jan 8 '19 at 19:53
  • $\begingroup$ I got confused and looked for the derivative to be continuous. $\endgroup$ – איתן לוי Jan 8 '19 at 20:02
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Your conclusion is right for $p>1$ and all values of $q>0$. The reason why you don't observe so on the graph is the the oscillation of the function increases around $x=0$ so it's indistinguishable to see whether the function is differentiable in $x=0$ or not. Also the function has no continuous derivative in $x=0$ for $0<p\le 1$. The figure below shows why:

enter image description here

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