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> [$\lim\_{n \to +\infty} n^{\frac{1}{n}} $](https://math.stackexchange.com/questions/115822/lim-n-to-infty-n-frac1n) 
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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?

Answer: 
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.
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, $$\lim\_{n\to\infty}n^{\frac{1}{n}}=1.$$