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

• Randall
In summary, the conversation discusses finding the limit as n goes to infinity of the quotient (ln(n+2))/(ln(2n)). One person suggests using L'Hospital's Rule while another suggests using the property ln(ab)=ln(a)+ln(b) to simplify the problem. Ultimately, the group agrees on using L'Hospital's Rule and the final answer is found to be 2. However, one person does not seem to fully understand the solution and there is further discussion about using Vela's hint.

## 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.

Randall said:

## 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

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Randall said:
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.

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 :(

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