What is the limit of $3^{1/n}$ when n approaches infinity Graphically, I see that $\lim_{n->\infty}3^{1/n}$ approaches $1$. However, how to show $\lim_{n->\infty}3^{1/n} = 1$ step by step? 

Binomial theorem:
For $x\ge 0:$
$(1+x)^n \ge 1 + nx + n(n-1)\dfrac{x^2}{2!} \ge $
$n^2\dfrac{x^2}{4}$ for $n\ge 2.$
Set $x=\dfrac{2√3}{n}:$
$(1+\dfrac{2√3}{n})^n \ge n^2\dfrac{(4)(3)}{4n^2}= 3;$
$ (1+\dfrac{2√3}{n})^{n} \ge 3$ , or
$1 +\dfrac{2√3}{n} \ge 3^{1/n} .$
With lower bound $1$:
$1\lt 3^{1/n} \le 1+ \dfrac{2√3}{n}.$
Limit $n \rightarrow \infty$ is?
