Is this a sufficient condition for non uniform continuity? Suppose $f$ is defined on $(-∞,∞)$. If the magnitude of its derivative $|f'(x)|$ approaches $∞$ as $x$ tends to $±∞$, then is it true that $f$ is not uniformly continuous on $(-∞,∞)$?
I have observed this for a number of functions. $f(x)=x$ is uniformly continuous on $(-∞,∞)$ whereas $g(x)=x^2$ is not.
 A: Yes. Consider $x<y$. By Lagrange's theorem, we have the estimate $$\lvert f(y)-f(x)\rvert=\lvert f'(\xi_{x,y})\rvert\cdot \lvert y-x\rvert\ge\lvert x-y\rvert\cdot\inf_{z\in [x,y]}\lvert f'(z)\rvert\ge \lvert x-y\rvert\cdot\inf_{z\in [x,\infty)}\lvert f'(z)\rvert$$
Now, assume as a contradiction the weaker hypothesis that for some $\varepsilon>0$ there is $\delta>0$ such that for all $x$, $\lvert f(x+\delta)-f(x)\rvert\le \varepsilon$. Then, for any fixed $x$, $$\delta\inf_{z\in [x,\infty)}\lvert f'(z)\rvert\le\lvert f(x+\delta)-f(x)\rvert\le\varepsilon$$
Taking the least upper bound over $x$, $$\delta\cdot\sup_{x\in\Bbb R}\inf_{z\in [x,\infty)} \lvert f'(z)\rvert\le\varepsilon$$
Or, since $\delta>0$ and $\sup\limits_{x\in\Bbb R}\inf\limits_{z\in [x,\infty)} \lvert f'(z)\rvert\in [0,\infty]$,
$$\sup_{x\in\Bbb R}\inf_{z\in [x,\infty)} \lvert f'(z)\rvert\le\frac\varepsilon\delta$$
But by defintion $\sup\limits_{x\in\Bbb R}\inf\limits_{z\in[x,\infty)}\lvert f'(z)\rvert=\liminf\limits_{x\to\infty}\lvert f'(x)\rvert=+\infty$, absurd.
Added 1: It must be noted that the hypothesis $\liminf\limits_{x\to\infty} \lvert f'(x)\rvert=+\infty$ is used in its fullest. By which I mean: there are strictly increasing differentiable functions such that $\limsup\limits_{x\to\infty} f'(x)=+\infty$ and $0<\liminf\limits_{x\to\infty} f'(x)<+\infty$.
Added 2: The proof above can be summarized in one inequality: if we call $$\Omega(f,\varepsilon):=\sup\left\{\delta\ge 0\,:\,\forall x\in\Bbb R,\ \sup_{0\le\theta<\delta} \lvert f(x+\theta)-f(x)\rvert\le\varepsilon\right\}$$ the continuity modulus of $f$ - so that $f$ is UC if and only if $\Omega(f,\varepsilon)>0$ for all $\varepsilon>0$ -, then $$\liminf_{x\to\infty}\lvert f'(x)\rvert\le\inf_{\varepsilon>0}\frac{\varepsilon}{\Omega(f,\varepsilon)}$$
