# Is the function $f(x)=x\sin(1/x)$ continuous and differentiable at x=0? [closed]

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.

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.

• 330587.In a nutshell.Very nice!+1. Feb 2 '18 at 14:29
• Very nice hint, that forces the OP to think by himself and do some self work. +1 Feb 2 '18 at 14:43

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 $$\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}$$

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 $x=0$.

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.

• Thank you for the feedback! Feb 2 '18 at 14:51