Explanation for divergence of $\ln(x)$ $\ln(x)$ diverges as $x \to \infty$, yet the differential $\frac{1}{x}$ tends to $0$ as $x \to \infty$. To the naive mind, it seems that if the differential tends to $0$ then the original function should asymptote and, therefore, converge. I appreciate that this is elementary, but an intuitive explanation would really help me.
 A: I'd guess your intuition doesn't have any problem with a strictly increasing continuous function $f\colon \mathbb R\to\mathbb R$ such that both $f(x)$ and $f'(x)$ tend to infinity as $x\to \infty$. Two examples of such functions are $x\mapsto \exp(x)$ and $x\mapsto x^3$. Since any such $f$ is injective, theres a left inverse $f^{-1}\colon \operatorname{im}(f) \to \mathbb R$. Now since $f'(x)$ tends to infinity, $(f^{-1})'(y)$ will tend to $0$ when $y\to\infty$. Hence the inverses of the examples I gave, namely $x\mapsto \ln(x)$ and $x\mapsto \sqrt[3]{x}$ will have the desired properties, tending to infinity with derivative tending to $0$.
I hope this explains why this behaviour isn't that unintuitive after all.

The relationship between $f'$ and $(f^{-1})'$ is the following: Since we have
$$
f^{-1}(f(x)) = x,
$$
the chain rule yields
$$
(f^{-1})'(f(x))\cdot f'(x) = 1.
$$
Hence, when $y=f(x)$, we have $(f^{-1})'(y) = \frac{1}{f'(x)}$. The intuition behind this is that the graph of $f^{-1}$ is the graph of $f$ reflected at the diagonal $y=x$. This transforms a slope $\frac{\Delta y}{\Delta x}$ into the reciprocal $\frac{\Delta x}{\Delta y}$.
Now for $x\to\infty$, we see that $(f^{-1})'(y) = \frac{1}{f'(x)} \to 0$, provided that $f'(x)\to\infty$.
