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$\newcommand{\sgn}{\operatorname{sgn}}$ So i have couple of questions about differentiability and continuity. For example lets consider $f(x)=\sin(\frac1x)$. It is defined and continuous for $x\neq0$ . It's derivative is $\frac{-\cos(\frac1x)}{x^2}$. It looks to me that derivative is also continuous for $x\neq0$. Is that correct?
If it is does, this means that $f(x)$ is defferentiable on $(-\infty, 0) \cup (0, + \infty)$? I know that $f(x)$ is not differentiable at $x=0$ because it is not defined there.

Let's look now at function $$ f(x) = \begin{cases} |x|^p\sin(\frac1x) & \text{for }x\ne0, \\ 0 & \text{for }x=0. \end{cases} $$

Here are three questions:

For which $p$ is $f(x)$ continuous? For which $p$ is $f(x)$ differentiable? For which $p$ is $f'(x)$ continuous?

Since $|x|^p\sin(\frac1x)$ for $x\neq0$ is product and composition of continous functions that are all defined for $x\neq0$, it follows that $f(x)$ is also continious for $x\neq0$.

Let $p=0$. In that case for $x\neq0$ i get $f(x)=\sin(\frac1x)$. In that case $f(x)$ discontinous in 0, because $\lim_{x\to 0}\sin(\frac1x)$ is oscilatinge beetween $[-1, 1]$ and $f(0) = 0$. In other words $\lim_{x\to 0}f(x)$ doesn't exist and it cant be fixed.

Let $p<0$. In that case for $x\neq0$ i get $f(x)=\frac{\sin(\frac1x)}{|x|^p}$. In that case $f(x)$ discontinous in 0, because $f(x)=\frac{\sin(\frac1x)}{|x|^p}$ is oscilatinge beetween $[-\infty,+ \infty]$ and $f(0) = 0$. In other words $\lim_{x\to 0}f(x)$ doesn't exist and it cant be fixed.

Let $p>0$. Since $f(x)$ = $|x|^p\sin(\frac1x)$ for $x\neq0$ and it is product and composition of continious functions for $x\neq0$, this means that $f(x)$ = $|x|^p\sin(\frac1x)$ is continious. Since $\lim_{x\to 0}|x|^p = 0$ and $\lim_{x\to 0}\sin(\frac1x) $ is oscilating between $[-1, 1]$, this means that $\lim_{x\to 0}f(x) = 0$. Also $f(0)=0$, this means that f(x) is continious on whole $R$.

Thus $f(x)$ is continuous for $p>0$.

Next question is: for which $p$ is $f(x)$ differentiable?

This is the question i am stuck on. I am assuming that it means differentiable on whole $R$ , if that's the case then answer is for $p>0$.

$$f'(x) = \begin{cases} { p|x|^{p-1}\sin(\frac1x)\sgn(x) - \frac{|x|^{p-1}\cos(\frac1x)}{x^2}}& \text{for }x\neq0, \\ 0& \text{for }x=0 \end{cases}$$

Third question is: For which $p$ is $f'(x)$ continuous?. I tried home and it seems to me that for $p\leq1$ $\lim_{x\to 0}{ p|x|^{p-1}\sin(\frac1x)\sgn(x) - \frac{|x|^{p-1}\cos(\frac1x)}{x^2}}$ doesn't exist . For $p>1$ i don't know how to compute the limit. For lets say $p=3/2$ i get $f'(x) = { p|x|^{\frac12}\sin(\frac1x)\sgn(x) - \frac{|x|^{\frac12}\cos(\frac1x)}{x^2}}$. While $p|x|^{\frac12}\sin(\frac1x)\sgn(x)$ goes to $0$ i dont know how to calculate $\frac{|x|^{\frac12}\cos(\frac1x)}{x^2}$, since it is undefined. Any help would be appreciated.

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Valiant but in some cases clumsy use of MathJax. I think I've clean it up, mostly. –  Michael Hardy Jan 1 at 18:16

1 Answer 1

Your discussion about the continuity properties looks acceptable (although I did not check every single statement you make).

I suggest that you compute directly $$ \lim_{x \to 0} \frac{f(x)-f(0)}{x} $$ to investigate the differentiability of $f$ at $x=0$. Your approach is not equivalent to the differentiability of the function: if $\lim_{x \to 0}f'(x)$ exists then $f'(0)$ exists too. But the opposite implication is, in general, wrong.

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