Are these 100 series enough?

[Dr.Peterson]

Are you suggesting [imath]\displaystyle \lim_{n \to \infty}\sqrt[\ln n]{\frac{1}{(\ln n)^{\ln n}}}[/imath]?

Can you please start your approach to see what I didn't see?

I'd start by considering the correct limit: [imath]\displaystyle \lim_{n \to \infty}\sqrt[n]{\frac{1}{(\ln n)^{\ln n}}}=\lim_{n \to \infty}\frac{1}{(\ln n)^{\frac{\ln n}{n}}}=\lim_{n \to \infty}(\ln n)^{-\frac{\ln n}{n}}[/imath]

If you were given this limit to solve, what would you do?

 

[blamocur]

I'd start by considering the correct limit: [imath]\displaystyle \lim_{n \to \infty}\sqrt[n]{\frac{1}{(\ln n)^{\ln n}}}=\lim_{n \to \infty}\frac{1}{(\ln n)^{\frac{\ln n}{n}}}=\lim_{n \to \infty}(\ln n)^{-\frac{\ln n}{n}}[/imath]

If you were given this limit to solve, what would you do?

Looks like the root limit is 1 here, doesn't it?

 

[Dr.Peterson]

Yes, I probably missed a step in my admittedly quick attempt; my main point here has been to recommend trying things.

 

[nasi112]

I'd start by considering the correct limit: [imath]\displaystyle \lim_{n \to \infty}\sqrt[n]{\frac{1}{(\ln n)^{\ln n}}}=\lim_{n \to \infty}\frac{1}{(\ln n)^{\frac{\ln n}{n}}}=\lim_{n \to \infty}(\ln n)^{-\frac{\ln n}{n}}[/imath]

If you were given this limit to solve, what would you do?

[imath]\displaystyle \lim_{n \to \infty}(\ln n)^{-\frac{\ln n}{n}} = \lim_{n \to \infty}e^{-\frac{\ln n}{n}(\ln n)}[/imath]

The power of e is [math]\frac{\infty}{\infty}[/math]
L'hoptital is applicable. I don't remember the derivative of [imath]-\ln n \ln n[/imath]. Do I take the product rule or I first combine [imath]-\ln n \ln n = -\ln^2 n[/imath]? Last time I did this type of derivative was 2 years ago.

 

[Dr.Peterson]

I don't remember the derivative of [imath]-\ln n \ln n[/imath]. Do I take the product rule or I first combine [imath]-\ln n \ln n = -\ln^2 n[/imath]? Last time I did this type of derivative was 2 years ago.

Either will work. One is probably a little less work.

In keeping with my theme here ... give both a try! That will help you learn which, if either, to try next time you see a similar problem. And if you tried an it failed ... it's just that much more experience.

On the other hand ... check your work! Your rewrite is wrong, so you'll get still more experience out of this if you take the derivatives of both the wrong expression, and then the right one.

 

[nasi112]

If my memory still telegraphing the product rule I get [imath]\displaystyle -\frac{1}{n}\ln n - \ln n \frac{1}{n}[/imath]. With the other rule I get [imath]\displaystyle -2\ln n \frac{1}{n}[/imath]. They are equal?

[imath]\displaystyle \lim_{n \to \infty}(\ln n)^{-\frac{\ln n}{n}} = \lim_{n \to \infty}e^{-\frac{\ln n}{n}(\ln n)} = \lim_{n \to \infty}e^{-\frac{2\ln n}{n}}[/imath]

The power of e is still giving [math]\frac{\infty}{\infty}[/math]
Do I repeat again the derivative?

 

[Dr.Peterson]

If my memory still telegraphing the product rule I get [imath]\displaystyle -\frac{1}{n}\ln n - \ln n \frac{1}{n}[/imath]. With the other rule I get [imath]\displaystyle -2\ln n \frac{1}{n}[/imath]. They are equal?

You tell me! Can you rewrite the first to look like the second?

[imath]\displaystyle \lim_{n \to \infty}(\ln n)^{-\frac{\ln n}{n}} = \lim_{n \to \infty}e^{-\frac{\ln n}{n}(\ln n)} = \lim_{n \to \infty}e^{-\frac{2\ln n}{n}}[/imath]

The power of e is still giving [math]\frac{\infty}{\infty}[/math]
Do I repeat again the derivative?

If you mean, apply L'Hopital's rule again, yes, that is often necessary.

But before you do that, go back and correct what I told you was wrong: As I said, your rewrite of the limit is incorrect. It is not true that [imath]\ln(x)=e^{\ln(x)}[/imath].

 

[nasi112]

I see the wrong.

[imath]\displaystyle \lim_{n \to \infty}(\ln n)^{-\frac{\ln n}{n}} = \lim_{n \to \infty}e^{-\frac{\ln n}{n}(\ln(\ln n))}[/imath]

With product rule I get [imath]\displaystyle -\frac{1}{n}\ln(\ln n) - \ln n \frac{1}{\ln n}\frac{1}{n}[/imath]

[imath]\displaystyle \lim_{n \to \infty}(\ln n)^{-\frac{\ln n}{n}} = \lim_{n \to \infty}e^{-\frac{\ln n}{n}(\ln(\ln n))} = \lim_{n \to \infty}e^{-\frac{\ln(\ln n) + 1}{n}}[/imath]

The power of e is giving [math]\frac{\infty}{\infty}[/math]
With product rule next time I get [imath]-\frac{1}{\ln n}\frac{1}{n}[/imath]

[imath]\displaystyle \lim_{n \to \infty}(\ln n)^{-\frac{\ln n}{n}} = \lim_{n \to \infty}e^{-\frac{\ln n}{n}(\ln(\ln n))} = \lim_{n \to \infty}e^{-\frac{\ln(\ln n) + 1}{n}} = \lim_{n \to \infty}e^{-\frac{1}{n\ln n}}[/imath]

The power of e is giving [math]\frac{1}{\infty} = 0[/math]
[imath]\displaystyle \lim_{n \to \infty}(\ln n)^{-\frac{\ln n}{n}} = \lim_{n \to \infty}e^{-\frac{\ln n}{n}(\ln(\ln n))} = \lim_{n \to \infty}e^{-\frac{\ln(\ln n) + 1}{n}} = \lim_{n \to \infty}e^{-\frac{1}{n\ln n}} = \lim_{n \to \infty}e^{-0} = 1[/imath]

Thanks Doctor to mention my mistake. I didn't see it in the beginning haha.