<s>[INST] Why is $ \lim_{x \to \infty} \ x^{2/x} = 1$ Why is $\displaystyle \lim_{x \to \infty} \ x^{2/x} = 1$ since this is an indeterminate form $\infty^{0}$ and I can't see any manipulation that would suggest this result?
 [/INST] $\lim \limits_{x \to \infty} x^{\frac{1}{x}} = 1$
Proof using AM-GM and Sandwich Theorem
$\frac{1 + 1 + 1 + \dots + \sqrt{x} + \sqrt{x}}{x} \geq \sqrt[x]{x} \geq 1$
$\frac{x - 2 + 2\sqrt{x}}{x} \geq \sqrt[x]{x} \geq 1$
$1 - \frac{2}{x} + \frac{2}{\sqrt{x}} \geq \sqrt[x]{x} \geq 1$
$\lim \limits_{x \to \infty} 1 - \frac{2}{x} + \frac{2}{\sqrt{x}} = 1$
$\therefore \lim \limits_{x \to \infty} x^{\frac{1}{x}} = 1 \implies \lim \limits_{x \to \infty} x^\frac{2}{x} = 1$
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