$\epsilon-\delta$ proof that $f(x) = x \sin(1/x)$ on $(0,1)$ is uniformly continuous I have to prove
$$
f: (0,1) \to [-1,1], \ \ f(x)=x\sin(1/x)
$$
using $\epsilon-\delta$.

Try
Fix $\epsilon >0$. I have
$$
\left|x \sin(1/x) - y \sin (1/y)\right| \le |\sin (1/x)| |x-y| + |y| |\sin (1/x) - \sin(1/y)|
$$
I can control $|\sin (1/x)| |x-y|$ part, since I can control $|x-y|$ and $|\sin(1/x)| \le 1$. But I don't know how to deal with $|y| |\sin (1/x) - \sin(1/y)|$ part.
Any help will be appreciated.
 A: Using the prosthaphaereis formula
$$|y| \left|\sin \frac{1}{x} - \sin\frac{1}{y} \right| = 2|y|\left|\sin \frac{x^{-1}-y^{-1}}{2} \right| \left|\cos \frac{x^{-1}+y^{-1}}{2} \right| \leqslant 2|y|\left|\frac{x^{-1} - y^{-1}}{2} \right| \\\leqslant \frac{|x-y|}{|x|}$$
WLOG assume $x > y$ -- otherwise, work with the alternative estimate $|x| \left|\sin \frac{1}{x} - \sin\frac{1}{y} \right|$.
Note that we also have 
$$|y| \left|\sin \frac{1}{x} - \sin\frac{1}{y} \right| \leqslant 2|y|$$
Suppose $|x - y| < \frac{\epsilon^2}{2}$.  Either $|y| < \frac{\epsilon}{2}$ and we have 
$$|y| \left|\sin \frac{1}{x} - \sin\frac{1}{y} \right| \leqslant 2\frac{\epsilon}{2} = \epsilon,$$ 
or 
$$|x| > |y| \geqslant \frac{\epsilon}{2},$$
and
$$|y| \left|\sin \frac{1}{x} - \sin\frac{1}{y} \right|\leqslant \frac{|x-y|}{|x|} < \frac{2}{\epsilon} \frac{\epsilon^2}{2} = \epsilon$$
Thus, $|x-y| < \frac{\epsilon^2}{2}$ implies $|y| \left|\sin \frac{1}{x} - \sin\frac{1}{y} \right|\leqslant \epsilon$ for all $x,y \in (0,1)$.
The estimate on the first line is only sharp enough to prove uniform continuity on an interval $[a,1)$ where $a > 0$.  However, we know that $x \sin \frac{1}{x} \to 0 $ as $x \to 0$ and this implies continuous extendibility to $[0,1)$ and, hence uniform continuity on $(0,1)$.  The argument above combines both of these elements into an $\epsilon - \delta$ proof.
