Why is $\lim_{x \to +\infty} \log(x) = +\infty$? I would have expected that the value of this limit is some fixed number, since
$$\frac{\mathrm d}{\mathrm dx} \log x = \frac1x$$
and
$$\lim\limits_{x \to +\infty} \frac1x = 0,$$
so the tangent of $\log(x)$ approaches $0$ if $x$ is large enough.
How do I prove that the limit of $\log x$ is indeed infinity?
Best Answer
(1) The fallacy in your heuristic argument: just because something is growing slowly, doesn't mean it doesn't get arbitrarily big. The $\log$ function is extremely slow growing (and it grows slower and slower as $x$ gets larger and larger, i.e. its second derivative is negative), but indeed its limit is infinity.
(2) To show that the limit is infinity, you just need to show that you can make the function as big as you want by taking $x$ large enough. More formally, you must show that for any large target $N$, there exists an $x_0$ such that $\log(x) > x_0$ whenever $x > N$.
Now $\log(x)$ is an increasing function (which you know already because you calculated its derivative, which is positive), so that means that whenever $x > e^N$, then $\log(x) > N$, because $\log(e^N) = N$ by definition. (I assume you mean the base $e$ logarithm because of the derivative calculated in your question, but if you meant some other base, of course just put that wherever you see an $e$.) This is then enough to conclude.