Limit of Root of Positive Real Number/Proof 2

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 $\sequence {x^{1/n} }$ is bounded below by $1$.

Now consider $x^{1/n} / x^{1 / \paren {n + 1} }$:

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

and so $\sequence {x^{1 / n} }$ is strictly decreasing.

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

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

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

So we take the subsequence:
 * $\sequence {x^{1 / {2 n} } }$

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

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