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$$f(x) = \begin{cases} x\sin(1/x), & \text{if $x$ $\ne$ $0$} \\ 0, & \text{if $x$ = $0$} \\ \end{cases}$$

Is $f$ continuous on $(-1/\pi$, 1/$\pi$)? Is $f$ differentiable on $(-1/\pi$, 1/$\pi$)?

I have a question with this problem. I know how to prove continuity on a single point, bu I'm not sure how to prove continuity for a whole interval. Also, I know there is a theorem that states that if a function is differentiable at a point, then it's continuous but I have a feeling that $f(x)$ is continuous but not differentiable.

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  • $\begingroup$ To prove continuity in internal go for critical point at x =0 because at other places function is xsin(1/x) which is continuous as it is product of two continuous function $\endgroup$ – DSinghvi May 9 '14 at 19:07
  • $\begingroup$ I remember this function when I was in High School, my professor gave this an example of function which is continuous but not differentiable. $\endgroup$ – Santosh Linkha May 9 '14 at 19:16
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To check continuity at $x=0$ use the squeeze lemma: clearly $ -1< \sin x < 1 \ \forall x$, so $\lim_{x \to 0} f(x) =0 = f(x_0)$, so the function is continuous. To check differentiability use the definition: $f'(0) = \lim_{h \to 0} \frac{f(0+h)-f(0)}{h}$. what do you get?

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By first principles $f'(0)=\lim_{h\to 0}\frac{f(h)-f(0)}{h}=\lim_{h\to 0}\frac{h\sin(1/h)}{h}=\lim_{h\to 0}\sin(1/h)$

Which is undefined, so yes it is not differentiable at $0$.

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Your feeling is true. To check $f$ is continuous on$(-1/\pi,1/\pi)$, you only need to check $\lim_{x\to 0}f(x)=0$. This can be proved using Squeeze_theorem.

For the undifferentiability, prove that $$ f^\prime(0)=\lim_{h\to 0}\frac{f(h)-f(0)}{h}=\lim_{h\to 0}\frac{h\sin(1/h)}{h}=\lim_{h\to 0}\sin(1/h) $$ doesn't exist.

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