My Math Forum limit to infinity of ln n / ln (n+1)

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 August 20th, 2009, 12:11 PM #1 Newbie   Joined: Jul 2009 Posts: 7 Thanks: 0 limit to infinity of ln n / ln (n+1) I'm trying to figure out how to find the limit of ln n / ln(n+1) as n approaches infinity. The answer booklet says it's 1. How do I approach this?
 August 20th, 2009, 12:32 PM #2 Senior Member   Joined: Mar 2009 Posts: 318 Thanks: 0 Are you allowed to use l"Hospital's Rule?
 August 20th, 2009, 12:55 PM #3 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: limit to infinity of ln n / ln (n+1) Use $\frac{\ln n}{\ln(n+1)}=\frac{\ln\left((n+1)\frac{n}{n+1}\rig ht)}{\ln(n+1)}=\frac{\ln(n+1)+\ln\left(\frac{n}{n+ 1}\right)}{\ln(n+1)}=1+\frac{\ln\left(\frac{n}{n+1 }\right)}{\ln(n+1)}.$
 August 20th, 2009, 07:12 PM #4 Newbie   Joined: Jul 2009 Posts: 7 Thanks: 0 Re: limit to infinity of ln n / ln (n+1) Thanks for the replies. Yes I looked at the rule I can use it. I just have another question what is the derivative of f(x) = ln (x+1)? Do I let u = x + 1, then multiply the derivative of f(x) with the derivative of u?
 August 20th, 2009, 11:19 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,262 Thanks: 1958 You multiply the derivative of f(u) with respect to u by the derivative of u with respect to x and then replace u with x + 1.
 August 21st, 2009, 06:59 AM #6 Newbie   Joined: Jul 2009 Posts: 7 Thanks: 0 Re: limit to infinity of ln n / ln (n+1) Once I use the rule I'll end up with (x + 1) / x. Do I just divide each literal by x?
 August 21st, 2009, 11:17 AM #7 Global Moderator   Joined: Dec 2006 Posts: 20,262 Thanks: 1958 Yes.

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# limit ln (n 1)

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