Limit of Root of Positive Real Number/Proof 1

Proof
Let us define $a_1 = a_2 = \cdots = a_{n-1} = 1$ and $a_n = x$.

Let $G_n$ be the geometric mean of $a_1, \ldots, a_n$.

Let $A_n$ be the arithmetic mean of $a_1, \ldots, a_n$.

From their definitions:
 * $G_n = x^{1/n}$

and:
 * $A_n = \dfrac {n - 1 + x} n = 1 + \dfrac{x - 1} n$

From Arithmetic Mean is Never Less than Geometric Mean:
 * $x^{1/n} \le 1 + \dfrac{x - 1} n$

That is:
 * $x^{1/n} - 1 \le \dfrac{x - 1} n$

There are two cases to consider: $x \ge 1$ and $0 < x < 1$.

Let $x \ge 1$.

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

Thus:
 * $0 \le x^{1/n} - 1 \le \dfrac 1 n \paren {x - 1}$

But from Sequence of Powers of Reciprocals is Null Sequence:
 * $\dfrac 1 n \to 0$ as $n \to \infty$

From the Combination Theorem for Sequences:
 * $\dfrac 1 n \paren {x - 1} \to 0$ as $n \to \infty$

Thus by the Squeeze Theorem:
 * $x^{1/n} - 1 \to 0$ as $n \to \infty$

Hence, again from the Combination Theorem for Sequences:
 * $x^{1/n} \to 1$ as $n \to \infty$

Now let $0 < x < 1$.

Then $x = \dfrac 1 y$ where $y > 1$.

But from the above:
 * $y^{1/n} \to 1$ as $n \to \infty$

Hence by the Combination Theorem for Sequences:
 * $x^{1/n} = \dfrac 1 {y^{1/n} } \to \dfrac 1 1 = 1$ as $n \to \infty$