is $$f(x) =\begin{cases} x^2\sin(\frac{1}{x}) \mbox{ for } x\neq 0 \\ 0 \mbox{ for } x= 0\end{cases}$$ a continuous function specially at point x=0?

And why being derivable its derivative is not continuous at x=0 since

$$f`(x) =\begin{cases} 2x\sin\frac{1}{x}-\cos\frac{1}{x} \mbox{ for } x\neq 0 \\ 0 \mbox{ for } x= 0\end{cases}$$?

I suspect this as to do with being derivable a function f $\mathbb{R}^n\rightarrow\mathbb{R}^m$ that don't imply its partial derivatives are continuous .this I don't get it so far.however partials derivatives if continuous is derivable.

Can you provide me a more general case than just one variable where f is derivable but its partials are not continuous?


$$0\leq \left|x^2\sin\left(\frac{1}{x}\right)\right|\leq x^2\to 0$$ if $x\to 0$. Then your function is continuous in $x=0$. Moreover, your function is derivable in $x=0$. Indeed,

$$\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0} x\sin\left(\frac{1}{x}\right)=0.$$

The fact that $f'$ is not continuous on $x=0$ implies just that f is not $\mathcal C^1(\mathbb R)$ not that $f$ is not derivable on $\mathbb R$. This is the perfect example of a function derivable with a derivate not continuous.

  • $\begingroup$ What if f was$$f(x) =\begin{cases} x^2\sin(\frac{1}{x})+5 \mbox{ for } x\neq 0 \\ 0 \mbox{ for } x= 0\end{cases}$$ is this function still continuous on x=0 or not since when x$\rightarrow$0 limf(x-)=f(x+)=5 but f(0)=0? @idm $\endgroup$ – studentNk Aug 27 '14 at 17:29
  • $\begingroup$ What makes f` not continuous on x=0 ? $\endgroup$ – studentNk Aug 27 '14 at 17:31
  • $\begingroup$ it would not be continuous in $x=0$ because $\lim_{x\to 0}f(x)=5\neq f(0)=0$ and so, it would not be derivable on $x=0$ (because not continuous at $x=0$) $\endgroup$ – idm Aug 27 '14 at 18:22

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