Is $x\sin\left(\frac{1}{x}\right)$ uniformly continuous in $\mathbb{R}$? Let $f:\mathbb{R}\xrightarrow{}\mathbb{R}$ with rule defined as
$$
f(x) = \left\{\begin{array}  xx\operatorname{sin}\left(\frac{1}{x}\right) &\text{  if }x\neq0 \\
                      0  &\text{ if } x=0
\end{array}\right.
$$
I would like to show that this function is uniformly continous over $\mathbb{R}$.
I started out with the $\epsilon-\delta$ definition and tried to manipulate the following expression without success,
$$
\left|x \operatorname{sin}\left(\frac{1}{x}\right)   - y \operatorname{sin}\left(\frac{1}{y}\right) \right |\\
=\left|x \operatorname{sin}\left(\frac{1}{x}\right) -x \operatorname{sin}\left(\frac{1}{y}\right) +x \operatorname{sin}\left(\frac{1}{y}\right)   - y \operatorname{sin}\left(\frac{1}{y}\right) \right | \\
\leq \left|x \left(\operatorname{sin}\left(\frac{1}{x}\right) - \operatorname{sin}\left(\frac{1}{y}\right)\right) + (x - y) \operatorname{sin}\left(\frac{1}{y}\right) \right |
$$
Also, the function isn't Lipschitz, so I can't go this way too.
I've seen some other proofs at mathSE however they try to show things for smallers domains as $[0,1]$.
 A: Hints:
(1) The function is continuous on any compact interval $[-a,a]$.
(2) The derivative is bounded on semi-infinite intervals $(-\infty,-a]$ and $[a,\infty)$ where $a > 0$.
A: Since $$\lim_{x\to\ 0} f(x) = f(0)$$
we know that $f$ is continuous particularly on $[-1,1]$. Hence, it is uniform continuous on $[-1,1]$.
Now, assume $x \in \mathbb R \setminus [-1,1]$. Then, by Lagrange's theorem (or MVT), we have
$$\exists \xi \in (x,y): f(x)-f(y) = f'(\xi)(x-y) \tag 1$$
Also,
$$|f'(x)| = \left|\sin\frac 1x - \frac 1x\cos \frac 1x\right| \le 1 + 1 = 2 \tag 2$$
Therefore,
$$(1)\land (2) \implies |f(x) - f(y)| = |f'(x)||x-y| \le 2|x-y|$$
Now, apply $\varepsilon$-$\delta$ definition.
A: hint
$ f $ is even and continuous at $ \Bbb R $ .
Given $ \epsilon>0$.
$$\lim_{|x|\to+\infty}f(x)=1$$
So, for $ |x| $ and $ |y| $ large enough,
Let us say, $$x>y>A>0\text{ or  } x<y<B<0$$
$$|f(x)-f(y)|\le|f(x)-1|+|f(y)-1|<\epsilon$$
On the other hand, $ f $ is uniformly continuous at the compact $ [A,B] $,
So, There exists $ \eta>0 $, such that
for $ x,y\in[A,B] $
$$|x-y|<\eta\implies |f(x)-f(y)|<\epsilon$$
Now, use continuity at $ A $ and $ B$ for the cases
$$x<A<y \text{ or } x<B<y$$
