Is $f(x) = x^3 \sin \frac{1}{x} $ uniformly continuous on $(0, \infty)$? Since the derivitive of $f$ is bounded on a neighborhood of $0$, $f$ is uniformly continuous on $(0, M)$ where $M$ is any positive number. I'd like to prove that $f$ is uniformly continuous on a neighborhood of $\infty$. The derivity of $f$ goes to $\infty$ as $x \rightarrow \infty$. So I guess $f$ is not uniforly continuous on $(0, \infty)$. I 'll appreciate it if you give me a proof. 
 A: $f$ is not uniformly continuous.
Short explanation:
For $x \to \infty$ we have asymptotically $\sin \frac 1x \approx \frac 1x$, so $f(x) \approx x^2$. 
A bit longer explanation: There is an $y_0 > 0$ such that $0<\sin(y) < 2y$ for $0<y<y_0$. Consequently, $\sin(\frac 1x) > \frac 1 {2x}$ and $f(x)>\frac {x^2}2$ for $x > \frac 1{y_0}$.
A function which is uniformly continuous can only grow linearly, though. For an $\delta, \epsilon$ instance of the definition one would have the bound $f(x) \leq f(0) + \frac \epsilon \delta x$.
A: It is not uniformly continuous on $(0,+\infty)$. We apply the following simple theorem, which immediately says the function not uniform continuous on $[1,+\infty)$

Theorem: Suppose $f(x)$ is uniformly continuous on $[1,+\infty)$, then $f(x)\over x$ is bounded on $[1,+\infty)$.
Proof:
By uniform coninuity, $\exists \delta  > 0$, such that when $\left| {x - y} \right| < \delta ,{\rm{ }}\left| {f(x) - f(y)} \right| < 1$.
Hence $$\begin{array}{l}
 - 1 < f(1 + \delta ) - f(1) < 1\\
 - 1 < f(1 + 2\delta ) - f(1 + \delta ) < 1\\
...\\
 - 1 < f(1 + n\delta ) - f(1 + (n - 1)\delta ) < 1
\end{array}$$
Therefore, we can prove that $\forall n \in \mathbb{N}, - n + f(1) < f(1 + n\delta ) < n + f(1)$. Hence for $n \in \mathbb{N},\frac{{f(1 + n\delta )}}{{1 + n\delta }}$ is bounded.
Now, $\forall x > 1$, there is an integer $N$ such that $1 + N\delta  \le x < 1 + (N + 1)\delta $, which implies $\left| {f(x) - f(1+N\delta )} \right| < 1$. This concludes the proof.
A: Take $$s_n=n+\frac{1}{n},\quad t_n=n,$$
then $\lim_{n\to\infty}(s_n-t_n)=0$, but
$$\lim_{n\to\infty}(f(s_n)-f(t_n))
=\lim_{n\to\infty}\left[\left(n+\frac{1}{n}\right)^3\sin\frac{1}{n+\frac{1}{n}}
-n^3\sin\frac1n\right]=2.$$
This implies $f$ is not uniform continuous on $(0,\infty)$.
