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?
On
Take any positive $\;M\in\Bbb R^+\;$ , then since $\;\log x\;$ is monotonic increasing and since we know what its inverse function is:
$$\log x>M\iff x>e^M$$
On
The derivative going to 0 tells you that the rate of change is constantly decelerating towards 0, but it still keeps going up "fast enough" to eventually hit any number. Remember, $\log x=y$ means $e^y =x$. So, if you give me any given number you want to hit in the positive reals in the range of the logarithm, $y$, all I need to do is plug in $e^y$. Thus, all real numbers are in the range, so its unbounded (especially since it's increasing)
On
Using FTC and $\log' x = \frac1x$ You can see that $$\lim_{x\to\infty} \log x = \lim_{x\to\infty} \int_1^x \frac1t\ \mathrm dt \ge \lim_{x\to\infty} \sum_{k=2}^{\lfloor x\rfloor} \frac1k = \infty$$ Because the harmonic series diverges.
The last step uses the fact that $\frac1t$ is monotonically decreasing on $(0,\infty)$ so $\int_n^{n+1} \frac1t \ \mathrm dt \ge \frac1{n+1}$ and $\frac 1t > 0$ so $$\int_1^x \frac1t \ \mathrm dt \ge \int_1^{\lfloor x \rfloor} \frac1t\ \mathrm dt$$
On
When $x\to +\infty$ we can make $x\ge e^K$ for any $K\in \mathbb{R}^+.$ $$x\ge e^K\iff\ln x\ge K,\,\,\,\,\,\forall K\in \mathbb{R}^+$$
On
Although I fell for this myself, and I'm not good at generating formal proofs, in retrospect:
let's assume we have a function that doesn't go to infinity. Then it would have to be finitely bounded from above, meaning you can find an impossible output value not reachable by any input value. But that isn't the case, no matter how large y gets, there is always an x that can produce this (obvious once one inverts the function to an exponential. IE take e^x=y, then there is no forbidden values for x). One can be mislead into thinking otherwise if the function or its inverse leans heavily towards one axis, but this is mathematically irrelevant.
also... if the tangent thus derivative goes to 0 at infinity, then the original function is definately bounded. so since 1/x is the derivative of log(x), and 1/x goes to 0 at infinity, then log(x) is bounded. But here's the clou: the horizontal line that bounds it.... can itself have a height of infinity, as is in this case.
So if a function is finitely bounded and monotonic, then its derivative definitely goes to 0 at infinity. If a function's derivative goes to 0 at infinity, it is definitely bounded, but can still go to infinity, because that bound can be infinity itself (ie it can be infinitely bounded), and every function is bounded by infinity anyways albeit in a weaker sense (if its derivative does not go to 0 at infinity).
(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.