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.

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

$\displaystyle \lim_{u \to 0} \ln((u+1)^{\frac{1}{u}}) = \lim_{n \to \infty}\ln((1+\frac{1}{n})^n)$ (letting u = 1/n)

$\displaystyle =\ln(\lim_{n \to \infty} (1+\frac{1}{n})^n) = \ln(e) = 1$

two comments:

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

2. the number e is often DEFINED as:

$\displaystyle \lim_{n \to \infty} (1 + \frac{1}{n})^n$