# continuity, uniformly continuous

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 $x \neq 0$? – Benjamin Dickman Nov 2 '12 at 21:20
Perhaps the OP meant to say "it is uniformly continuous on a bounded subset of $\mathbb{R}$"? – Valentin Nov 2 '12 at 22:30
this is a similar question where the idea may be borrowed math.stackexchange.com/questions/223147/simple-calculus-inquiry/… – Valentin Nov 2 '12 at 23:06

Hint I'll prove continuity at $x=0$. To prove it is continuous at zero, you need to prove $\lim_{x\to 0} f(x)=f(0)$.

You are given $f(0)=0 \,.$ You need to prove the limit of the function is $0$ too. Now, notice this,

$$\left |x^2\sin\left(\frac{1}{x}\right) \right|\leq |x^2|=x^2 \implies x^2\sin\left(\frac{1}{x}\right) \to 0 \quad \mathrm{as}\,\,\, x\to 0 \,,$$ By the sandwich theorem. Note that the fact that $|\sin(x)|\leq 1$ has been used.

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Mhenni has already proved continuity. To prove$f$ is uniformly continuous on any bounded subset of $\mathbb{R}$, use the fact that a continuous function on a compact set is uniformly continuous on that set. Note that a bounded set in $\mathbb{R}$ is contained in a bounded interval, which is compact.

To prove that $f$ is uniformly continuous on all of $\mathbb{R}$: note that $\displaystyle\lim_{x\to\infty} x^2 \sin(1/x) = \lim_{|x|\to\infty} x \frac{\sin(1/x)}{1/x} = x$ and $x$ is a uniformly continuous function. You can formalize this relatively easily.

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You can observe that $|f^{'}(x)|<2$ is bounded fore every x in $\mathbb{R}$ and so $f(x)$ is a lipschitz function and then it is uniformly continuous. $$f^{'}(x)= 2*x*sin(1/x)+x^2*cos(1/x)*(-1/x^2)$$ – Agenog Apr 9 '13 at 13:02
Also we note that $f^{'}(x)$ has the point of discontinuity of the second species in x = 0 (in fact, the derivative may have only types of discontinuity of the second species) – Agenog Apr 9 '13 at 13:16