Show that $x\ln(x)$ is not uniformly continuous on $(0,\infty)$ I am trying to show that $f(x) = x\ln(x)$ is not uniformly continuous on the interval $(0,\infty)$. The solution given here Show $f(x)=x\ln x$ is not uniformly continuous  does it by using $\epsilon-\delta$ but I want to do it by sequences if possible. The "x" term is messing things up because I cannot take $$x_n = \textrm e^{-n } $$ and $$y_n = \textrm e^{-n + 1}$$ because $|f(x) -f(y)|$ does not work out. Any ideas on sequences I can take?
 A: Take $x_n = e^{n^2}$ and $y_n = x_n + 1/n$. Note that $|x_n - y_n| \to 0$. 
By the MVT there exists $\xi_n \in (x_n ,x_n + 1/n)$ such that
$$y_n \ln y_n - x_n \ln x_n = (1 + \ln \xi_n)/n > (1+\ln x_n)/n > n \to +\infty$$
More generally 
A differentiable function $f:(0,\infty) \to \mathbb{R}$ is not uniformly continuous if $|f'(x)| \to +\infty$ as $x \to +\infty$. (It is not enough for $f'$ to be unbounded.)
For any $\delta > 0$ take $x \in (0,\infty)$ and $y = x + \delta/2$.  We have $|x-y| < \delta$, but there exists $\xi \in (x,y)$ such that
$$|f(x) - f(y)| = |f'(\xi)||y-x| = |f'(\xi)|\delta/2$$  If $|f'(x)| \to +\infty$ then there exists $X$ such that for all $\xi > x  > X$ we have $|f'(\xi)| > 2/\delta$ and, hence $|f(x) - f(y)| > 1$.
So for any $\delta > 0$ there exists $x,y \in (0,\infty)$ with $|x - y| < \delta$ and $|f(x) - f(y) > 1$, and $f$ is not uniformly continuous.
A: Let $x_n=n$ and $y_n=n+\frac 1 {\log n}$. Note that $\bigl(n+\frac 1 {\log n}\bigr) \log \bigl(n+\frac 1 {\log n}\bigr)-n \log n \geq \frac 1 {\log n} (\log n) \geq 1$.
