How to prove that $\lim\limits_{n\to\infty}\sqrt[n]{p}=1$? Could you tell me how to show if $p>0$ then$\lim\limits_{n\to\infty}\sqrt[n]{p}=1$?
(+clues)
1.put $\sqrt[n]{p}=1+h_{n}$
2.Bernoulli's inequality
If you don't mind, use the clues to prove it.
 A: I'm sure there's a more in-depth version, but the simple answer is: $\lim\limits_{n\to\infty}\sqrt[n]{p}= \lim\limits_{n\to\infty}p^{\frac{1}{n}} = p^0$
A: Assume $p>1$ then by Bernoulli
$$1+nh_n\leq (1+h_n)^n=p$$
and so
$$h_n\leq \frac{p-1}{n}$$ It follows that $h_n \to 0$.
A: There are two cases:
$0 < p < 1$
and
$p > 1$.
Here is the second:
Let $p^{1/n} = 1+a$.
Then
$p
=(1+a)^n
\ge 1+an
\gt an
$
so
$a < p/n$
so
$a \to 0$
as
$n \to \infty$.
If
$0 < p < 1$,
let
$p = \dfrac1{1+a}
$
and do a similar thing.
As often,
nothing original here.
As a matter of fact,
this question
is a duplicate.
A: For $p>1$ let $p^{1/n}=1+h_n.$ Then $ h_n>0.$ We show that $h_n\to 0$ as $n\to \infty.$ For $n\geq 2$  we have $$p=(1+h_n)^n=1+h_n\binom {n}{1}+(h_n)^2\binom {n}{2}+...\;>(h_n)^2\binom {n}{2}\implies$$  $$\implies  p>(h_n)^2\binom {n}{2}=(h_n)^2 n (n-1)/2\implies$$  $$\implies 2 p/n(n-1)>(h_n)^2\implies$$  $$\implies 2\sqrt p /(n-1)>\sqrt {2 p/n(n-1)}\;>h_n>0.$$ 
For $0<p<1$ let  $p'=1/p.$ Then $p'>1$ so $( p')^{1/n}\to 1,$ so $p^{1/n}=1/(p')^{1/n}\to 1.$ The case $ p=1$ is trivial.
A: When $n$ goes to infinity (by the way it likes) $1/n$ goes to $0$. When $\alpha$ goes to $0$ (by the way it likes) $p^{\alpha}$ goes to $1$ (because $p^{\alpha}$ is continuous). That is it.
And you don't have to use algebra at all.
