Jamir Summers

2022-11-25

How to prove that $\underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{n}=1$ ?

Dillan Foley

Expert

Rewrite as
$\sqrt[n]{n}={n}^{1/n}={e}^{\mathrm{ln}\left(n\right)/n}$
Now, you can write that
$\underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{n}=\underset{n\to \mathrm{\infty }}{lim}{e}^{\mathrm{ln}\left(n\right)/n}={e}^{\underset{n\to \mathrm{\infty }}{lim}\mathrm{ln}\left(n\right)/n}$
Looking at the exponent, you have (using L'Hopital's Rule)
$\underset{n\to \mathrm{\infty }}{lim}\frac{\mathrm{ln}\left(n\right)}{n}=\underset{n\to \mathrm{\infty }}{lim}\frac{1/n}{1}=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}=0$
Therefore, you have
$\underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{n}={e}^{0}=1$

DinamisGr

Expert

Observe that
$\sqrt[n]{n}={n}^{1/n}={e}^{\frac{1}{n}\mathrm{log}\left(n\right)}.$
Since $e:\mathbb{R}\to \mathbb{R}$ is a continuous function, taking the limit gives
$\underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{n}=\underset{n\to \mathrm{\infty }}{lim}{e}^{\frac{1}{n}\mathrm{log}\left(n\right)}={e}^{\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}\mathrm{log}\left(n\right)}.$
It can be shown in a variety of ways (Taylor expansion comes to mind)
$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}\mathrm{log}\left(n\right)=0,$
and hence
$\underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{n}={e}^{0}=1.$

Do you have a similar question?