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I am trying to understand the continuity and differentiability of this function: $f(x) = \begin{cases} x\sin(\frac{1}{x}) & \text{if $x \ne 0$}\\ 0 & \text{if $x = 0$} \end{cases}$

If any indeterminate form arises in the calculation, I like to know how to deal with it and how to explain it.

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Hint

It's indeed continuous but not derivable. For continuity : What is $$\lim_{x\to 0}x\sin(1/x)\ \ ?$$ For non derivability, show that $$\lim_{x\to 0}\sin(1/x)$$ doesn't exist.

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  • $\begingroup$ 330587.In a nutshell.Very nice!+1. $\endgroup$ Feb 2 '18 at 14:29
  • $\begingroup$ Very nice hint, that forces the OP to think by himself and do some self work. +1 $\endgroup$
    – DonAntonio
    Feb 2 '18 at 14:43
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Try using limits. For continuity, consider the limit $$\lim_{x\to 0}x\sin\left(\frac{1}{x}\right)=\lim_{u\to\infty}\frac{\sin(u)}{u}$$For the derivative, you should consider the definition, i.e. consider the limit $$\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}$$

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Since $\displaystyle\left|x\sin\frac{1}{x}\right|\le|x|$, $\displaystyle\lim_{x\to0}f(x)=f(0)$. $f$ is continous at $x=0$.

$\displaystyle \lim_{x\to 0}\frac{f(x)-f(0)}{x}=\lim_{x\to 0}\sin\frac{1}{x}$ does not exist. $f$ is not derivable at $x=0$.

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It is obviously continuous since$$-x\le x\sin\dfrac{1}{x}\le x$$but not differentiable because$$\lim_{x\to 0}\dfrac{f(x)}{x}=\lim_{x\to 0}\dfrac{x\sin\dfrac{1}{x}}{x}=\lim_{x\to 0}\sin\dfrac{1}{x}$$which doesn't exist so doesn't the derivation.

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  • $\begingroup$ Thank you for the feedback! $\endgroup$ Feb 2 '18 at 14:51

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