Uniform continuity of $f(x) = x\cos(x)$ Is it uniformly continuous? $ f: [0,\infty]  $
$$f= x\cos(x)$$ 
I proved that if 
$$|x_1 -x_2| \rightarrow 0 \quad
\mathrm{then}\quad  |f(x_{1})-f(x_{2})| \rightarrow 1 $$
so it is not uniformly continuous, right?
 A: Assume the function is uniformly continuous. This means that for a fixed $\varepsilon > 0$ we can find $\delta > 0$ so that:
$$
\forall x_1, x_2 \in [0, \infty) : |x_1 - x_2| < \delta \implies \left|x_1 \cos(x_1) - x_2 \cos(x_2)\right| < \varepsilon
$$
Pick any $x_1$ and $x_2$ that satisfy $|x_1 - x_2| < \delta$. We can add $2\pi n$ to both and maintain that $|x_1 - x_2| < \delta$. However:
\begin{align*}
\left|f(x_1 + 2\pi n) - f(x_2 + 2\pi n)\right| &= \left|(x_1 + 2\pi n)\cos(x_1 + 2\pi n) - (x_2 + 2\pi n)\cos(x_2 + 2\pi n)\right| \\
&= \left|\left(f(x_1) - f(x_2)\right) + 2\pi n\left(\cos(x_1) - \cos(x_2)\right)\right| \\
&\ge 2\pi n\left|\cos(x_1) - \cos(x_2)\right| - \varepsilon
\end{align*}
By making $n$ as large as necessary, we can make $\left|f(x_1 + 2\pi n) - f(x_2 + 2\pi n)\right| > \varepsilon$. Hence, $f$ is not uniformly continuous.
A: If $f'(x)$ were bounded you could conclude that $f$ is uniformly continuous. Since $f'(x)$ is unbounded for $x\to\infty$ you cannot conclude anything from looking merely at the derivative. Note that $x\mapsto\sqrt{x}$ is uniformly continuous on ${\mathbb R}_{>0}$, but its derivative is unbounded.
In order to prove that the given $f$ is not uniformly continuous on ${\mathbb R}_{>0}$ we have to produce an $\epsilon_0>0$ and point pairs $x$, $y>0$ arbitrarily close to each other with $|f(y)-f(x)|\geq\epsilon_0$.
Choose $\epsilon_0:=1$ and put
$$x_n:=2n\pi+{3\pi\over2}\,\qquad y_n:=x_n+{\pi\over n}\qquad(n\geq3)\ .$$
Then $y_n-x_n={\pi\over n}$. Furthermore there is a $\xi\in[x_n,y_n]$ with $$f(y_n)-f(x_n)={\pi\over n}f'(\xi)={\pi\over n}\bigl(\cos\xi-\xi\sin\xi\bigr)\geq{\pi\over n}\>\xi\>{1\over2}\geq\pi^2>\epsilon_0\ .$$
A: We can employ the exact same idea as in this answer. Take $x_n=2n^2\pi+\pi/2+1/n$ and $y_n=(2n^2+\pi/2)$. Then $\lim\limits_{n\to\infty}|x_n-y_n|=0,$ however, $$\begin{aligned}|f(x_n)-f(y_n)|&=|(2n^2\pi+\pi/2+1/n)\underbrace{\cos(2n^2\pi+\pi/2+1/n)}_{=-\sin(1/n)}-(2n^2\pi+\pi/2)\underbrace{\cos(2n^2\pi+\pi/2)|}_{=0}\\&=(2n^2\pi+\pi/2+1/n)\sin(1/n)\\&>2n^2\pi\sin(1/n)\\&=2n\pi\frac{\sin(1/n)}{1/n}\to+\infty\end{aligned}$$
Take $\varepsilon=1$. For sufficiently large $n\in\Bbb N,$ the last expression exceeds it and beyond no matter how small $\delta>0$ is, $|x_n-y_n|<\delta$ is never sufficient. Adding $\pi/2$ just boiled down the problem to the one of proving $f:\Bbb R\to\Bbb R, f(x)=x\sin(x)$ isn't uniformly continuous on $\Bbb R.$
