While checking convergence of integrals, I always have to remind myself that $\log x$ goes to infinity very slowly. This is almost like and axiom to me. I don't intuitively get why it goes to infinity so slowly. On a graph, it sure doesn't look that way. Any thoughts?
More precisely: $$\forall\delta>0: \lim_0x^\delta\log x=0$$
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$\log(x)$ is monotone increasing (we can check this because $(\log x)' = \frac{1}{x}>0$ for all $x>0$). So, it has two options:
It's bounded above, so $\log(x)\to L$, it's bound.
It's unbounded, so $\log(x)\to\infty$.
Now, assume it's bounded by something huge, like $10^{1000}$. This doesn't have to be the lowest/best bound, just a bound. But, we see that $\log(x)>10^{1000}$ for all $x> e^{10^{1000}}$. This is true for any bound $L$ we can choose, $\log(x)>L$ for $x>e^L$.
When we say that it grows slowly, it means that for $\log(x) > 1000$, we need to have $x > e^{1000}$. As $e^2\approx 9$, we get that $e^{1000}\approx 10^{500}$ (this is a very rough estimate). So, for $\log x$ to be greater than $1000$ (which, when talking about a function going to infinity, is "small"), we need to have $x$ to be approximately $10^{500}$, which is "large". This should give you an idea of the "scale" of how slowly it grows.
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Intuitively speaking, the expression $\log_e x$ is asking the question, $e$ to the what power gives $x$? Since exponential growth is very fast, the answer to this question doesn't need to be very big to achieve $x$, so you could imagine that this expression doesn't grow very quickly.
On
Another way to see it is to observe that logarithms transform exponents in multiplication.
For example, if you go to infinity with a sequence $(x_{n})_{n}$ defined for each $n$ by $x_{n}=2^{n}$, you have a very fast convergence. Indeed, $2^{n}$ goes very very fast to infinity.
But if you look at $\lim_{x\to\infty}\log(x)$, you know by continuity of the logarithm that:
$$\lim_{x\to\infty}\log(x)=\lim_{n\to\infty}\log(x_{n})$$
for any sequence $x_{n}$ going to infinity when $n$ goes to infinity. Now, recall the particular sequence $x_{n}=2^n$ I took. It is a very fast convergence to infinity. However, we know by the property of the logarithm that:
\begin{align*} \lim_{x\to\infty}\log(x)&=\lim_{n\to\infty}\log(x_{n})\\ &=\lim_{n\to\infty}\log(2^{n})\\ &=\lim_{n\to\infty}n\log(2)\\ &=\underbrace{\log(2)}_{\text{constant }<1 }\lim_{n\to\infty}n \end{align*}
And this convergence is way slower than $2^{n}$. This means that if a quantity goes to infinity exponentially, the logarithm of this quantity goes linearly to infinity, which is super-slow in comparison to exponential growth.
Of course, these properties are consequences of the fact the logarithm is the reciprocal of the exponential, but I think it is clearer to see in terms of sequences.
You can think of this this way.
The function $x\mapsto \log (x)$ goes to $-\infty$ when $x$ goes to $0$ as slow as the function $x\mapsto e^x$ goes to $0$ fast when $x$ goes to $-\infty$.
This is because $\log(e^x)=x$.
And then, think that $x\mapsto e^x$ goes to $0$ when $x$ goes to $-\infty$ as fast as it goes to $+\infty$ when $x$ goes to $+\infty$.
Which is really fast, faster for instance than any $x\mapsto \vert x^n\vert $ as you can imagine.