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

I know that 
$$\lim_{n\rightarrow\infty}\sqrt[n]n=1$$
and I can imagine that $n$ grows linearly while $n$th root compresses it exponentially and therefore the result is $1$, but how do I calculate it?
 A: $$\lim_{n\rightarrow\infty}\sqrt[n]n$$
$$=\lim_{n\rightarrow\infty}e^{\frac{\ln(n)}{n}}$$
and as we know that $\lim_{n\rightarrow\infty}\frac{\ln(n)}{n} = 0$ {apply  l'Hospital's rule}
So $$\lim_{n\rightarrow\infty}\sqrt[n]n=1$$
A: Claim: For each $a>1$, there exists $N$ such that $n<a^n$ for all $n>N$.
Proof: Write $a=1+b$.  By the binomial theorem, $a^n=(1+b)^n\geq \frac{1}{2}n(n-1)b^2$ when $n\geq 2$.  Thus $\frac{a^n}{n} \geq \frac{1}{2}(n-1)b^2$.  It follows that if $N$ is at least $2/b^2+1$, then when $n>N$, $n<a^n$.

As a consequence, for each $a>1$, there exists $N$ such that $1\leq n^{1/n}<a$ for all $n>N$, and this implies that $\lim\limits_{n\to\infty}n^{1/n}=1$.
A: One could use the fact that for a sequence of positive terms, if $\lim\limits_{n\rightarrow\infty}{a_{n+1}\over a_n}$ exists, then so does $\lim\limits_{n\rightarrow\infty} \root n\of {a_n}$ and the two limits are equal. A proof of this general fact can be found in these notes of Pete L. Clark. This result can also be found in many analysis texts; e.g., baby Rudin.
That your sequence has limit $1$ is easily shown using the above fact. A detailed proof that your sequence has limit $1$, based on the proof of the above fact,  can be found in this thread.
A: Let's see a very elementary proof. Without loss of generality we proceed replacing $n$ by $2^n$ and get that:
$$ 1\leq\lim_{n\rightarrow\infty} n^{\frac{1}{n}}=\lim_{n\rightarrow\infty} {2^n}^{\frac{1}{{2}^{n}}}=\lim_{n\rightarrow\infty} {2}^{\frac{n}{{2}^{n}}}\leq\lim_{n\rightarrow\infty} {2}^{\frac{n}{\dbinom{n}{2}}}=2^0=1$$
By Squeeze Theorem the proof is complete.
A: Let$$y=n^{\frac{1}{n}}$$ 
$$\log y=\log( n^{\frac{1}{n}})$$
$$\Rightarrow \lim_{n\to\infty}\log y=\lim_{n\to\infty}\frac{\log n}{n}$$
We have $\left(\frac{\infty}{\infty}\right)$ form so we apply L'hospital rule, we get,
$$ \lim_{n\to\infty}\log y=\lim_{n\to\infty}\frac{\frac{1}{n}}{1}$$
$$\therefore \lim_{n\to\infty}\log y=\lim_{n\to\infty}\frac{1}{n}=0$$
Hence
$$\lim_{n\to\infty} y=e^{0}=1.$$
It follows that, $$\lim_{n\to\infty}n^{\frac{1}{n}}=1.$$
A: Perhaps one of the most elementary ways to prove it: since $\,n\geq 1\,\,\forall n\in\mathbb{N}\,$ , we can put$$\sqrt[n]{n}=1+c_n\,,\,c_n\geq0\Longrightarrow n=(1+c_n)^n\geq \frac{n(n-1)}{2}c_n^2$$using the binomial expansion ,so that$$0<c_n\leq\sqrt\frac{2}{n-1}$$and now just apply the squeeze theorem and get $\,c_n\to 0\,$, which is precisely what we need.
