Show that: $f:\mathbb{R} \rightarrow \mathbb{R}$ where $f(x) = x^2 + x$ is not uniformly continuous.

To prove that, I need to show that there are $x,y \in \mathbb{R}$ such that $\exists \epsilon > 0$, $\forall \delta > 0$ and it follows that $|x-y| \leq \delta$ but $|f(x) - f(y)| > \epsilon$.

First suppose that thee following statements are true:

• $x > \frac{1}{2\delta}$
• $\epsilon = 1$
• $y = x + \delta$

Now let $\delta > 0$.

Clearly $|x-y| \leq \delta$ because $|x-y|=|x-(x+\delta)| = |\delta| = \delta \leq \delta$. Now see that $$ \begin{align*} |f(x)-f(y)|&=|x^2 + x - (x+\delta)^2 - (x+\delta)|\\ &=|2x\delta + \delta^2 + \delta|\\ &=2x\delta + \delta^2 + \delta\\ &>2x\delta = 2 > 1 = \epsilon \end{align*} $$

Terefore $f(x)$ is no uniformly continuous.

Can someone please check my proof? Also, it's a proved theorem that if $a_n,b_n$ are functions in the domain of $f$ such that $(a_n - b_n) \rightarrow 0$ and $f(a_n) - f(b_n) \nrightarrow 0$ then $f$ is not uniformly continuous. Can someone please show me an example of $a_n$ and $b_n$ for that case?

I tried to prove some not uniformly continuous functions using the previous theorem but it's not trivial (at least for me), to find those sequences...

Thanks!