Derivative when function approaches infinity. Let $f$ be a continous and differentiable function in it's natural domain.
If $$\lim_{x\to a-}f(x)=\infty$$, is it always true, that $$ \lim_{x\to a-} f'(x) \ge0 $$ if the limit exists (or is infinite)?
Intuitively this seems to be true, but can this be proven exactly?
 A: Define $L=\lim_{x\to a^-}f'(x)$, assuming this limit exists (or is infinite).
Suppose $L=-\infty$. Plug in any number $B$ to the definition of this limit, to get an interval $(a-\delta,a)$ within which $f'(x)$ is bounded above:
$$\exists B:\exists\delta>0:\forall x\in(a-\delta,a):f'(x)<B.$$
Suppose $L$ is finite. Plug in any number $\varepsilon>0$ to the definition of limit, and define $B=L+\varepsilon$, to get an interval $(a-\delta,a)$ within which $f'(x)$ is bounded:
$$\exists B>L:\exists\delta>0:\forall x\in(a-\delta,a):f'(x)\in(L-\varepsilon,L+\varepsilon)=(2L-B,B).$$
So, in any case with $L<+\infty$, we have $f'(x)<B$ for all $x$ close enough to $a$. (If $B$ is negative, then $f'(x)$ is also bounded by any positive number; so we can assume $B>0$.)
Now take any point $x$ within that interval. By the mean value theorem, $\frac{f(x)-f(a-\delta)}{x-(a-\delta)}$ is the derivative of $f$ at some point in that interval, so
$$\frac{f(x)-f(a-\delta)}{x-(a-\delta)}<B.$$
Note that $a-\delta<x<a$ implies
$$0<x-(a-\delta)=(x-a)+\delta<\delta,$$
so we have
$$f(x)-f(a-\delta)<B\cdot\big(x-a+\delta\big)<B\delta$$
$$f(x)<f(a-\delta)+B\delta.$$
This says that $f$ is bounded in that interval, which contradicts
$$\lim_{x\to a^-}f(x)=+\infty.$$
Therefore $L\not<+\infty$; either $L=\lim_{x\to a^-}f'(x)=+\infty$ or the limit doesn't exist.

Here's an example of a case where $f\to\infty$ but the limit of $f'$ doesn't exist:
$$f(x)=\frac{1}{x^2}-5\sin\frac{1}{x^2}$$
(graph).
A: Assume that $a$ is allowed to be infinity. Then, in general, the answer is no. For instance, let $f(x)= \ln x$. Then:
$$ \lim_{x \rightarrow \infty} \ln x = \infty$$
but:
$$ \lim_{x \rightarrow \infty} \frac{1}{x} = 0 $$
