# Limit at Infinity of x^n

From ProofWiki

## 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

From Upper Bound of Natural Logarithm:

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

which, by Combination Theorem for Continuous Functions, implies:

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \lim_{x \to +\infty} n \ln x\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle n \lim_{x \to +\infty} \ln x\) | \(\displaystyle \) | \(\displaystyle \) |

From Logarithm Tends to Infinity:

- $n \ln x \to +\infty$ as $x \to +\infty$

The result follows from the Push Theorem.

$\blacksquare$