# What is the lim as n goes to infinity of (ln(n+2))/(ln(2n))

## Homework Statement

what is the lim as n goes to infinity of (ln(n+2))/(ln(2n)) ?

## The Attempt at a Solution

It looks like you would get "small" infinity over "large" infinity, so does that make it 1? undefined? 0? thanks. Is there some simplifying I should be doing? Thanks.

berkeman
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## Homework Statement

what is the lim as n goes to infinity of (ln(n+2))/(ln(2n)) ?

## The Attempt at a Solution

It looks like you would get "small" infinity over "large" infinity, so does that make it 1? undefined? 0? thanks. Is there some simplifying I should be doing? Thanks.

Are you familiar with L'Hospital's Rule? http://en.wikipedia.org/wiki/L'Hôpital's_rule

Last edited:
vela
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It looks like you would get "small" infinity over "large" infinity, so does that make it 1? undefined? 0? thanks. Is there some simplifying I should be doing? Thanks.
##\frac \infty\infty## is an indeterminate form. To figure out if the limit exists and what it's equal to, you have to do some sort of manipulation. A common way to deal with an indeterminate form is to use, as Berkeman suggested, L'Hopital's rule. For this problem, you could also use the property ##\ln ab = \ln a + \ln b## to rewrite the quotient as
$$\frac{\ln (n+2)}{\ln 2n} = \frac{\ln [n(1+\frac 2n)]}{\ln 2n} = \frac{\ln n + \ln (1+\frac 2n)}{\ln n + \ln 2},$$ which you should be able to find the limit of now. Regardless of the method, you should get the same answer.

• berkeman
Thank you all - I use L'Hospital's rule and found the limit to be 2.

berkeman
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And what do you find using Vela's hint? :-)

I'm not sure I understood Vela's hint. That seems to have it more complex :(

vela
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I'm kinda guessing you didn't make much of an effort at understanding what I did, or if you did, there's some basic algebra you need to brush up on.

Anyway, can you show us your work? You didn't get the right answer.

• berkeman