The $n^{th}$ root of $n$

For the sake of a quick, "pure" math post.

It is a well known fact that $\lim_{n \to \infty} \sqrt[n]{n} = 1$

There are multiple proofs of that. One such proof is to use the inequality of Arithmetic and Geometric means

$$\frac{x_1 + x_2 + \dots + x_n}{n} \ge \sqrt[n]{x_1\cdot x_2\cdots x_n} \quad (x_i \ge 0)$$

Choose $x_1 = x_2 = \dots = x_{n-2} = 1$ and $x_{n-1} = x_n = \sqrt{n}$

Then we get,

$$\frac{n-2 + 2\sqrt{n}}{n} \ge \sqrt[n]{n} \ge 1$$

And so,

$$1 - \frac{2}{n} + \frac{2}{\sqrt{n}} \ge \sqrt[n]{n} \ge 1$$

By the sandwich theorem, the limit is one.