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I am trying to show that $f(x) = xln(x)$ is not uniformly continuous on the interval $(0,infty)$. The solution given here Show $f(x)=xln x$ 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?

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.

Let $x_n=n$ and $y_n=n+frac 1 {log, n}$. Note that $(n+frac 1 {log, n}) log (n+frac 1 {log, n})-n log , n geq frac 1 {log, n} (log, n) geq 1$.

Any differentiable function $f$
for which
$f'(x)$ is unbounded
is not uniformly continuous.

$(xln(x))'
=1+ln(x)
$
is unbounded as
$x to infty$
so it is not
uniformly continuous.