2
$\begingroup$

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: $$\forall\delta>0: \lim_0x^\delta\log x=0$$

  • 2
    Slowly compare to what??? – 2017-01-05
  • 1
    You must digest what means the logarithmic function. Then you can understand why it goes to infinity as slow. A logarithm is the inverse of exponentiation, start thinking with powers of base $10$, by example. – 2017-01-05
  • 1
    @barakmanos see edit! – 2017-01-05

5 Answers 5

3

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.

  • 1
    $e^x$ goes to zero slowly as $x\to-\infty$? On the contrary, it goes very fast, faster than any $|x^{-n}|$ for example... – 2017-01-05
  • 0
    @Rahul This is exactly how my confusion arose. – 2017-01-05
  • 0
    @Rahul Oups my mistake, thank you for correcting me. – 2017-01-05
  • 0
    @Michael: One could argue instead that $\log$ and $\exp$ are inverses, so one is fast when the other is slow and vice versa. For example $x^2$ is fast so $\sqrt x$ is slow. – 2017-01-05
  • 0
    @Rahul Absolutely, this is the point I am trying to make... – 2017-01-05
1

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.

0

$\log(x)$ is monotone increasing (we can check this because $(\log x)' = \frac{1}{x}>0$ for all $x>0$). So, it has two options:

  1. It's bounded above, so $\log(x)\to L$, it's bound.

  2. 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.

0

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.

0

Because $\log(x^{\delta})=\delta\log(x)$, which is essentially $\log(x)$, and

$$\log(\log(x))\leftrightarrow \log(x)\leftrightarrow x\leftrightarrow e^x\leftrightarrow e^{e^x}$$

truly represent different growth behaviors.