# Limit at Infinity of x^n

## Theorem

Let $x \mapsto x^n$, $n \in \R$ be a real function which is continuous on the open interval $\left({1 \,.\,.\, +\infty}\right)$.

If $n > 0$, then $x^n \to +\infty$ as $x \to +\infty$.

## Proof

$\forall n > 0: n \ln x < x^n$

which, by Combination Theorem for Continuous Functions, implies:

 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \lim_{x \to +\infty} n \ln x$$ $$=$$ $$\displaystyle$$  $$\displaystyle$$ $$\displaystyle n \lim_{x \to +\infty} \ln x$$ $$\displaystyle$$ $$\displaystyle$$
$n \ln x \to +\infty$ as $x \to +\infty$

The result follows from the Push Theorem.

$\blacksquare$