# Limit of ln(n)/ln(n+1) as n->+infinity, very confusing

1. Oct 26, 2008

### fiziksfun

Limit of ln(n)/ln(n+1) as n-->+infinity, very confusing

can someone help me find the lim as n approaches infinity of

ln(n)/ln(n+1)

I used L'HOP so it became (1/n)/(1/n+1) -- as this approaches infinity, it's 0/0, and this confuses me. What am I doing wrong?

2. Oct 26, 2008

### gabbagabbahey

Re: L'hop

$\frac{d}{dn} ln(n+1)=\frac{1}{n+1} \neq \frac{1}{n}+1$

Does that help ? ;0)

3. Oct 27, 2008

### boombaby

Re: L'hop

first, $$\frac{\frac{1}{n}}{\frac{1}{1+n}}=\frac{1+n}{n}=1+\frac{1}{n}$$ which approaches to 1, as n approaches to infinity...
second, I'm assuming n stands for integers. And L'hospitals Rule is not allowed to apply to a sequence.

Last edited: Oct 27, 2008
4. Oct 27, 2008

### HallsofIvy

Staff Emeritus
Re: L'hop

L'Hopitals Rule certainly is allowed to apply to a sequence! If f(x) has limit L as x goes to infinity then f(n) has limit L as n goes to infinity. This method is perfectly valid. Of course, you have to do the derivative and the algbra correctly!

5. Oct 27, 2008

### boombaby

Re: L'hop

yeah, I mean this exactly, but applying L'hospitals by brute force is not proper (derivative can be applyed to x but not to n).