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I was wondering if you could help me with this: $$ f(x)=\begin{cases} x^2\sin\left(\frac{1}{x}\right) & \text{ for } x \ne 0\\ 0 & \text{ for } x=0 \end{cases}. $$ I need to observe that f is continuous on $\mathbb{R}$ and then explain why it is uniformly continuous bounded subset of $\mathbb{R}$.

Finally, is $f$ uniformly continuous on $\mathbb{R}$? Do I take $f'(x)$?

I know that I should be using the theorem for f being continuous, which says that f is continuous for some $x_0$ and then evaluating it for the entire $\text{dom}(f)$.

Apologizes for the format of my post!

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    What do you mean by "explain why it is uniformly continuous bounded subset of $\mathbb{R}$"? It sounds like you're supposed to prove the range of this function is bounded, which is not the case. Perhaps you meant to write $f(x) = x \sin(\frac{1}{x})$ for $x \neq 0$?2012-11-02
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    Perhaps the OP meant to say "it is uniformly continuous *on a* bounded subset of $\mathbb{R}$"?2012-11-02
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    this is a similar question where the idea may be borrowed http://math.stackexchange.com/questions/223147/simple-calculus-inquiry/223191#2231912012-11-02

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