(letting u = 1/n)

two comments:

1. we can take the limit inside the ln function, since ln is continuous

2. the number e is often DEFINED as:

Results 1 to 2 of 2

- Nov 10th 2011, 09:07 PM #1

- Joined
- Sep 2009
- Posts
- 162

## why is it that ln(1+u)^(1/u)=1

why is it that as the limit of u approaches zero of ln(u+1)^(1/u)=1? I am trying to figure this out because I want to see the proof of the derivative of lnx = 1/x. I know that ln(u+1)^(1/u)=1 and when I plug in digits in my calculator I see it to be true, however, I cannot see it algebraically why it is true. I cannot see it because I see ln(1)^1/0. To me this is not possible.

- Nov 10th 2011, 10:04 PM #2

- Joined
- Mar 2011
- From
- Tejas
- Posts
- 3,546
- Thanks
- 842