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? 
 A: For any  number $a>0$, $a^{1/n}$ tends to $1$ when $n$ tends to infinity since 
$$\log a^{1/n}=\frac1n\,\log a\to 0\enspace\text{when }\enspace n\to\infty.$$
A: Note that
$$\lim_{n->\infty}\frac1n = 0 \quad \quad 3^0=1$$
thus by algebraic rule and continuity
$$\lim_{n->\infty}3^{1/n}=3^{\lim_{n->\infty}{1/n}}$$
we have
$$3^{1/n}\to 3^0=1$$
A: You only need two elementary results: The binomial theorem and the squeeze theorem. 
The binomial theorem implies that if $x>0$ then $(1+x)^n\geq 1+nx$.
Letting $x_n=3^{1/n}-1$, then $$(1+x_n)^{n}=3.$$ Since $x_n>0$, we have, by the above result, that $$3=(1+x_n)^n \geq 1+nx_n.$$
So $0\leq x_n\leq\frac{2}{n}$. Hence $x_n\to 0$ by the squeeze theorem, and and hence $3^{1/n}=x_n+1\to 1.$

This works for any sequence $a^{1/n}$ with $a>1$ since $x_n=a^{1/n}-1$ gives us that $0<x_n\leq \frac{a-1}{n}.$ 
To show it for $0<a<1$, 
we just need that $f(x)=\frac{1}{x}$ is continuous at $x=1.$ Assuming that, we use that $$ a^{1/n}=f\left(\left(a^{-1}\right)^{1/n}\right)$$ And if $0<a<1$, then $1<a^{-1}$ and by our previous result: $$\left(a^{-1}\right)^{1/n}\to 1.$$ So $a^{1/n}\to f(1)=1.$
A: HINT:
Using $e^x\le \frac1{1-x}$ for $x<1$, we have for $n\ge 2$
$$1< 3^{1/n}< \frac{1}{1-\frac1n \log(3)}$$
Now apply the squeeze theorem.
A: 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?
A: As an alternative, with a more advanced approach by the ratio-root criteria 
$$a_n = \sqrt[n]{b_n} \quad b_n=3$$
$$\frac{b_{n+1}}{b_n} \rightarrow L\implies a_n=b_n^{\frac{1}{n}} \rightarrow L$$
thus since
$$\frac{b_{n+1}}{b_n}=\frac33=1 \implies a_n = \sqrt[n]{3}\to 1$$
A: Bernoulli's Inequality, which, for integer exponents, can be proven using a simple inductive argument, says
$$
\left(1+\frac2n\right)^n\ge3\ge1
$$
Taking roots, we get
$$
1\le3^{1/n}\le1+\frac2n
$$
Then, the Squeeze Theorem ensures that
$$
\lim_{n\to\infty}3^{1/n}=1
$$
