1
$\begingroup$

When trying to evaluate$$\ln(\ln(\ln(\ln(\cdots\ln(x)\cdots))))$$I noticed that the answer was bound to be complex for any $x$.

Plugging in a very, very large real number in for $x$ will eventually become complex, and the $\log$ of a complex number is always complex or imaginary.

But what happens when we try to evaluate $$\lim_{x\to\infty}\ln(\ln(\ln(\ln(\cdots\ln(x)\cdots))))$$

We see that $$\lim_{x\to\infty}\ln(x)=\infty$$Meaning that if we try to do this repeatedly, we should still get infinite as our final result.

But we also see that for arbitrarily large $x$ where $x$ is real and finite, the result is a complex answer.

Which creates a sort of contradiction. What does it evaluate to?!?!

$\endgroup$
5
$\begingroup$

There is no contradiction. You are looking at two different limits.

Your first limit increases the number of evaluations of the $\ln$ function, a kind of tetration. We could invent a new notation and call this

$$\ln\uparrow_n(x)$$

where $n$ is the number of evaluations of the $\ln$ function. As you note,

$$\lim_{n\to\infty}\ln\uparrow_n(x)$$

for a fixed $x$ does not exist.

The other limit is for changing $x$. As you note,

$$\lim_{x\to\infty}\ln\uparrow_n(x)=\infty$$

for a fixed $n$. Those are different, since different variables are used in the limit. Of course, the double limit

$$\lim_{x\to\infty,\ n\to\infty}\ln\uparrow_n(x)$$

does not exist. There is nothing unusual here, since it is common for double limits to not exist, even when one or both single limits exist. This kind of thing is covered in just about any second-year Calculus class.

$\endgroup$
  • $\begingroup$ "Of course, the double limit does not exist" This is not obvious to me. How did you show that the double limit does not exist? $\endgroup$ – Will Sherwood Jan 2 '17 at 20:38
  • $\begingroup$ @WillSherwood: A theorem that is commonly covered in second-year calculus is that, if a double limit exists, then each single limit, when the other variable is constant, exists. The contrapositive is then: if one of the single limits does not exist then the double limit cannot exist. That's what we have here--the single limit for changing $n$ and fixed $x$ does not exist, so the double limit also does not exist. Of course, to understand this you need to know multivariable calculus and the meaning of a double limit, which is not the same as two single limits. Clear? $\endgroup$ – Rory Daulton Jan 2 '17 at 20:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.