Prove or Disprove: If $a_n$ was positive and its limit is positive then the limit of the $\sqrt[n]{a_n}$ is $1$ If $a_n$ was positive and its limit is positive then the limit of the $\sqrt[n]{a_n}$ is $1$. I think it has to do with multiplication to start here,but I couldnt figure it out.
If $a_n \to a &gt;0$, then $\log a_n \to \log a$ by continuity of log (and $\log a$ is finite). Also, $\frac{1}{n} \to 0$, so $\frac{1}{n} \log a_n \to (0) \log a = 0$. By continuity of exponentiation, we have $(a_n)^{1/n} = e^{\frac{1}{n} \log a_n } \to e^0 = 1$, as desired.
Here is an approach. Assuming $\lim_{n\to \infty}a_n = L&gt;0 $ and recalling the fact


  $$ \lim_{n\to \infty} (a_n)^{1/n} = \lim_{n\to \infty} \frac{a_{n+1}}{a_n} = \frac{L}{L} = 1.$$