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I have the function

$$f(x)= \begin{cases} x^2 \sin\left(\displaystyle\frac{1}{x}\right) & \text{if } x\ne 0 \\ 0 & \text{if }x=0 \end{cases} $$

and I have to show that it is continuous at $x_0=0$. I think they want me to use epsilon delta argument. Can somebody help?

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    Just put $\delta=\min {\{\epsilon, 1}\}$ – 2017-01-23

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I don't think they want $\delta-\epsilon$ proof. Im pretty sure that they want to proof the existence of $\lim_{x\to 0} f(x)$ and after show that $$\lim_{x\to 0} f(x)=0=f(0)$$ Aftter that you can conclude the continuity for the function!

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    ... and proving the existence of that limit is an $\epsilon$-$\delta$ proof. It might be camouflaged as a squeeze theorem, or anything else, but somewhere deep down there is an $\epsilon$-$\delta$ no matter what you do. – 2017-01-23
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    @Arthur If you look at it that way, *everything* proving continuity is an $\epsilon-\delta$ proof. – 2017-01-23
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    I know it. But also this type of exersise usually its been made to teach about the Lateral Limits. – 2017-01-23
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    @5xum I know, kind of like any proof of a statement like "For any natural number $n$, ..." is induction deep down. – 2017-01-23
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Hint: use the squeeze method between $-x^2$ and $x^2$.

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Hint:since $|\sin t|\le|t|$ for every t. See that $$|f(x)|\le|x|$$