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**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!