Find the limit of $u_n=\sqrt[n]{n}$ 
Possible Duplicate:
$\lim_{n \to +\infty} n^{\frac{1}{n}} $
Limit of the sequence $\lim_{n\rightarrow\infty}\sqrt[n]n$ 

Find the limit of $u_n=\sqrt[n]{n}$
I try to prove $\sqrt[n]{1}<\sqrt[n]{n}<$something that some thing <
 A: Since $\sqrt[n]{n} > 1$, write $\sqrt[n]{n} = 1 + a_n$ where $a_n > 0$. Thus:
$$
n = (1 + a_n)^n
$$
Apply the binomial theorem to the RHS. Can you use the above to squeeze $a_n$ and prove it converges? 
A: This is a classical sequence. It's an example in any instructional first course of analysis. There are several ways to obtain the limit of this sequence. But first of all we must prove that this sequence converge.
For all $\epsilon>0$ exist $n_\epsilon\in\mathbb{N}$ such that $n>n_\epsilon$ implies $n\cdot \epsilon< e^{n\cdot \epsilon+1}$. 
\begin{align}
n\cdot \epsilon < e^{n\cdot \epsilon+1}\implies 
&
\ln (n\cdot \epsilon) < n\cdot \epsilon+1
\\
\implies
&
\frac{1}{n}\ln (n\cdot \epsilon) < \frac{1}{n}(n\cdot \epsilon+1)
\\
\implies
&
\ln\big(\sqrt[n]{n}\cdot \sqrt[n]{\epsilon}\big) < 1\cdot \epsilon+\frac{1}{n}
\\
\implies
&
\ln\sqrt[n]{n}+ \ln\sqrt[n]{\epsilon} < 1\cdot \epsilon+\frac{1}{n}
\\
\implies
&
\ln\sqrt[n]{n} < 1\cdot \epsilon+\frac{1}{n}-\ln\sqrt[n]{\epsilon}
\\
\implies
&
\sqrt[n]{n} < e^{(1\cdot \epsilon+\frac{1}{n}-\ln\sqrt[n]{\epsilon})}
\\
\end{align}
Then 
$$
1< \sqrt[n]{n} < e^{(1\cdot \epsilon+\frac{1}{n}-\ln\sqrt[n]{\epsilon})}
$$
And  $\displaystyle\lim_{n\to\infty}\Big(1\cdot \epsilon+\frac{1}{n}-\ln\sqrt[n]{\epsilon}\Big)=\epsilon$ implies that for all $\epsilon>0$ exist $n_\epsilon\in\mathbb{N}$ such that $n>n_\epsilon$ we have
$$
1< \sqrt[n]{n} < e^{ \epsilon}.
$$
Then we have the result: $\lim_{n\to 0}{\sqrt[n]{n}}=1$
A: Its a lot simpler actually. Since $log(x)$ is a continuous function you have that $a_n\rightarrow a$ implies $log(a_n)\rightarrow log (a)$ whenever logs are defined. Try using this.
