Prove $\lim\limits_{n \to \infty }\sqrt[n]{a}=1$, if $a>0$ I tried solving by using $\log$ and got $\log(a)/n = \log (1)$ which after applying limit (of $n \to \infty$) gives $0= \log(1)$. Is this right?

Write , $a^{\frac{1}{n}} = e^{\ln(a^{\frac{1}{n}} )}= e^{\frac{1}{n} \cdot \ln(a) } $
Now , just apply the limit.
