# Limit of zero to the zero

• Aug 21st 2012, 05:58 AM
VinceW
Limit of zero to the zero
$\displaystyle \lim\limits_{n \to \infty} \left(\frac{1}{n}\right)^\frac{1}{n}$. The answer is 1, but how do I show this?

My textbook says for this scenario take $\displaystyle y = f(x)^{g(x)}$, take $\displaystyle \ln y$, find the limit of that, and then the limit of $\displaystyle y = e^{\ln y}$. I'm not sure how to do this one either: $\displaystyle \lim\limits_{n \to \infty} \ln \left[\left(\frac{1}{n}\right)^\frac{1}{n}\right]$

EDIT: OK solved.

$\displaystyle \lim\limits_{n \to \infty} \ln \left[\left(\frac{1}{n}\right)^\frac{1}{n}\right]$ $\displaystyle = \lim\limits_{n \to \infty} \frac{\ln \frac{1}{n}}{n}$ which is solved with l'hopital's rule, comes to zero, and the total limit comes to one. thanks!
• Aug 25th 2012, 07:01 PM
topsquark
Re: Limit of zero to the zero
Quote:

Originally Posted by VinceW
$\displaystyle \lim\limits_{n \to \infty} \left(\frac{1}{n}\right)^\frac{1}{n}$. The answer is 1, but how do I show this?

My textbook says for this scenario take $\displaystyle y = f(x)^{g(x)}$, take $\displaystyle \ln y$, find the limit of that, and then the limit of $\displaystyle y = e^{\ln y}$. I'm not sure how to do this one either: $\displaystyle \lim\limits_{n \to \infty} \ln \left[\left(\frac{1}{n}\right)^\frac{1}{n}\right]$

EDIT: OK solved.

$\displaystyle \lim\limits_{n \to \infty} \ln \left[\left(\frac{1}{n}\right)^\frac{1}{n}\right]$ $\displaystyle = \lim\limits_{n \to \infty} \frac{\ln \frac{1}{n}}{n}$ which is solved with l'hopital's rule, comes to zero, and the total limit comes to one. thanks!

There is as much material to suggest that $\displaystyle 0^0 = 0$. You have done the limit correctly (as far as I can see) but the problem $\displaystyle 0^0$ has been labelled as undefined.

-Dan
• Aug 25th 2012, 10:52 PM
JJacquelin
Re: Limit of zero to the zero
Hi VinceW !

Just have a look at this paper and you will see several commented examples of 0^0 :
"Zéro puissance zéro - Zero to the zero power"
Scribd