Show a function is not uniformly continuous given its derivative is unbounded, and they are both non-decreasing. quite strangely, my brain just seems to not understand how to formally write up the answer to this question.
Given $f:[0,\infty) \to\mathbb{R}$, $f'(x)$ is of order no less than n, e.g. $\frac{f'(n)}{n}$ not going to 0, and both $f$ and $f'$ are non-decreasing. Then $f$ is not uniformly continuous.
I think the reasoning is somewhat intuitive in the sense that if the derivative can blow up, then it can create a discontinuity in the equation itself. Also, the question should be addressed using the mean value theorem.
 A: It suffices to find $x_k$, so that $|x_k -n_k|\to 0$ ($n_k$ is a subsequence of $1, 2, 3, \cdots $) and
$$|f(x_k) - f(n_k)| \ge 1.$$
First of all, since $f$ is nondecreasing, $f\ge 0$. Thus the $f'(n)/n \ge 0$. If we assume that $f'(n)/n$ does not go to zero, there is $\epsilon_0>0$ and a subsequence $n_k$ so that 
$$f'(n_k)/n_k \ge \epsilon_0 \Rightarrow f'(n_k) \ge \epsilon_0 n_k.$$
Since $f$ is differentiable, by the mean value theorem, for all $y > n_k$, 
$$f(y) - f(n_k) = f'(y_k) (y-n_k), \ \ \ \text{for some } y\in (n_k, y).$$
Thus we have 
$$|f(y) - f(n_k)| = f(y) - f(n_k) \ge f'(n_k) (y-n_k) \ge \epsilon n_k (y-n_k).$$
Now choose $x_k = n_k + \frac{1}{\epsilon n_k}$. Then the condition are satisfied and thus $f$ is not uniformly continuous. 
Remark The condition on $f'(n)/n$ alone does not imply the result. There are differentiable function $g$ so that $g = 0$ outside $[0,1]$ and has unbounded derivatives. Then 
$$f = \sum_{n=1}^\infty g(x-k_n -2n)$$
for some suitably choosen $k_n$ will have $f'(n)/n\to +\infty$ and $f$ is uniformly continuous. 
