How to prove that $\displaystyle \lim_{n \to \infty} \sqrt[n]{n}=1$? I have seen this fact thrown around a lot and never really stopped to prove it; plugging in a few values convinced me of its truth. But I would like to see the result proved. To be clear, this is not homework, just human curiosity; I'm looking for any nice proof. Thanks.
 A: Rewrite as
$$
\sqrt[n]{n}=n^{1/n} = e^{\ln(n)/n}
$$
Now, you can write that
$$
\lim_{n\to\infty} \sqrt[n]{n} = \lim_{n\to\infty} e^{\ln(n)/n} = e^{\lim_{n\to\infty}\ln(n)/n}
$$
Looking at the exponent, you have (using L'Hopital's Rule)
$$
\lim_{n\to\infty}\frac{\ln(n)}n = \lim_{n\to\infty} \frac{1/n}{1}=\lim_{n\to\infty} \frac1n = 0
$$
Therefore, you have
$$
\lim_{n\to\infty} \sqrt[n]{n} = e^0 = 1
$$
A: Let $\sqrt[n]{n}=1+t$. Then $$n=(1+t)^n\geq\frac{n(n-1)}{2}t^2\Rightarrow t^2\leq\frac{2}{n-1}\rightarrow0~(n\rightarrow\infty)$$So $\lim_{n\rightarrow0}t=0$, for otherwise $\lim_{n\rightarrow\infty}t^2\neq0$, a contradiction.
A: You could use the fact that for a sequence $(a_n)$ 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.
This is easy to apply here ...
See these notes of Pete L. Clark for a proof of the above stated fact.
A: Observe that 
$$
\sqrt[n]{n} = n^{1/n} = e^{\frac{1}{n}\log(n)}.
$$
Since $e : \mathbb{R} \to \mathbb{R}$ is a continuous function, taking the limit gives 
$$
\lim_{n\to\infty}\sqrt[n]{n} = \lim_{n \to \infty}e^{\frac{1}{n}\log(n)} = e^{\lim_{n \to \infty} \frac{1}{n}\log(n)}.
$$
It can be shown in a variety of ways (Taylor expansion comes to mind) 
$$
\lim_{n\to\infty}\frac{1}{n}\log(n) = 0,
$$
and hence 
$$
\lim_{n\to\infty}\sqrt[n]{n} = e^0 = 1.
$$
