Limit of Root of Positive Real Number/Proof 2

Theorem
Let $x \in \R: x > 0$ be a real number.

Let $\left \langle {x_n} \right \rangle$ be the sequence in $\R$ defined as $x_n = x^{1/n}$.

Then $x_n \to 1$ as $n \to \infty$.

Proof
We consider the case where $x \ge 1$; when $0 < x < 1$ the proof can be completed as for proof 1.

From Root of Number Greater than One‎:
 * $x^{1/n} \ge 1$

Hence $\left \langle {x^{1/n}} \right \rangle$ is bounded below by $1$.

Now consider $x^{1/n} / x^{1/\left({n+1}\right)}$:

So:
 * $x^{1/n} > x^{\frac 1 {n+1}}$

and so $\left \langle {x^{1/n}} \right \rangle$ is strictly decreasing.

Hence from the Monotone Convergence Theorem (Real Analysis), it follows that $\left \langle {x^{1/n}} \right \rangle$ converges to a limit $l$ and that $l \ge 1$.

Now, since we know that $\left \langle {x^{1/n}} \right \rangle$ is convergent, we can apply Limit of Subsequence equals Limit of Sequence.

That is, any subsequence of $\left \langle {x^{1/n}} \right \rangle$ must also converge to $l$.

So we take the subsequence:
 * $\left \langle {x^{1/{2n}}} \right \rangle$

From what has just been shown:
 * $x^{1/{2n}} \to l$ as $n \to \infty$

Using the Combination Theorem for Sequences, we have:
 * $x^{1/n} = x^{1/{2n}} \cdot x^{1/{2n}} \to l \cdot l = l^2$ as $n \to \infty$

But a Convergent Real Sequence has Unique Limit, so $l^2 = l$ and so $l = 0$ or $l = 1$.

But $l \ge 1$ and so $l = 1$.

Hence the result.