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:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \lim_{x \to +\infty} n \ln x\) | \(=\) | \(\displaystyle \) | \(\displaystyle n\lim_{x \to +\infty} \ln x\) | \(\displaystyle \) | \(\displaystyle \) |
From Logarithm Tends to Infinity:
- $n \ln x \to +\infty$ as $x \to +\infty$
The result follows from Function Larger than Divergent Function is Divergent and Logarithm Tends to Infinity.
$\blacksquare$
