A weird limit involving e and 1/n

Hi, I'm working on a problem but stuck at a weird looking limit:

$\displaystyle \lim _{n \rightarrow \infty } [1- e^{-a-log(n)}]^2 $

It is suppose to converge to $\displaystyle e^{-e^{-a}}$ but:

I wrote the limit as $\displaystyle \lim _{n \rightarrow \infty } [1- \frac {1}{n}- \frac {1}{e^a}]^n $

and as $\displaystyle =\lim _{n \rightarrow \infty } [1 + \frac {-1-ne^{-a}}{n}]^n $

But neither are useful as I know the rule is $\displaystyle \lim _{n \rightarrow \infty } (1+ \frac {a}{n} )^n = e^a $ and I can't seem to get them down to something like that.

Any hints? Thank you!

Re: A weird limit involving e and 1/n

$\displaystyle e^{-a-\ln n}=e^{-a}\cdot e^{-\ln n}=\displaystyle \frac{1}{e^a}\cdot\frac{1}{n}=\frac{1}{ne^a}$

Re: A weird limit involving e and 1/n

Hey tttcomrader.

Hint: e^(-a-log(n)) = e^(-a)/n. Check the limit of lim n-> infinity [1 + b/n]^n (should be e^b if memory serves me correctly).