Proof that $x^2+2x-5$ is not uniformly continuous on all real numbers I understand that $x^2$ is not uniformly continuous on all real numbers, and have seen some good proofs for that, but specifically for the function
$f(x) = x^2 +2x - 5$
I am having trouble. I am trying a proof by contradiction.
My proof so far:
Suppose $f$ is uniformly continuous on all real numbers.
Let $\epsilon$ = 1. Since $\epsilon > 0$ and $f$ is uniformly continuous on all real numbers, there exists $\delta > 0$ such that for all $x$ and $x_0$ that are elements of all reals,
choose $x$ = ?
choose $x_0$ = ?
Then $|f(x) - f(x_0)|$ = |$(x^2 + 2x - 5) - (x_0^2 + 2x_0 - 5)$|
= |$(x^2 - x_0^2) + 2(x - x_0)$| = |$(x - x_0)(2 + x + x_0)$|
From here I don't know where to go, I just know that I somehow have to show that this is greater than 1 to contradict $|f(x) - f(x_0)| < \epsilon = 1$.

Let $\delta>0$. Take a real number $x$. Then $\left\lvert\left(x+\frac\delta2\right)-x\right\rvert=\frac\delta2<\delta$. But$$f\left(x+\frac\delta2\right)-f(x)=\frac\delta2f'(c),\tag1$$for some $c\in\left[x,x+\frac\delta2\right]$. So, take $x$ so large that $f'(x)>\frac2\delta$. Then $f'(c)>\frac2\delta$ and it will follow from $(1)$ that $f\left(x+\frac\delta2\right)-f(x)>1=\varepsilon$.
