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Consider the function $f\colon\mathbb R\to\mathbb R$ where $f$ is defined by $$f(x)= \begin{cases} x^b\sin(1/x), &\text{if $x>0$};\\ 0,&\text{if $x<=0$.} \end{cases}$$ Prove that the derivative $f'(x)$ is continuous at $x=0$ if and only if $b>2$.

Im not exactly sure what to do here, im thinking maybe something with left and right limits?

Any help would be appreciated!

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  • $\begingroup$ Is $b \in \mathbb{N}$? $\endgroup$ – Keba Jan 23 '14 at 0:40
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See that
if $b = 0$ then f is not even continous. if $b=1$ f is continuous but not differentible. if $b=2$ f is differentiable and derivative is not continuous. if $b=3$ f has continuous derivatives but the second derivative does not exists. if $b=4$ f is twice differentiabl but the second derivative is not continuous. and so on . . .

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For $\;x\neq 0\;$:

$$f'(x)=bx^{b-1}\sin\frac1x-x^{b-2}\cos\frac1x$$

For $\;x=0\;$:

$$f'(0)\stackrel?=\lim_{x\to 0}\frac{x^b\sin\frac1x}x=\lim_{x\to 0}x^{b-1}\sin\frac1x=\begin{cases}0&,\;\;b>1\\{}\\\text{doesn't exist}&\,\;\;b\le 1\end{cases}$$

Well, for $\;f'(x)\;$ to be continuous at zero it must be that $\;f'(x)\xrightarrow[x\to 0]{}f'(0)=0\;$ . For this to happen at least it must be that $\;f'(0)\;$ exists and thus $\;b>1\;$, but also$\;\ldots\ldots$

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