Limit at Infinity of x^n

From ProofWiki
Jump to: navigation, search


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 $x \to +\infty$.


From Upper Bound of Natural Logarithm:

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

which, by Combination Theorem for Continuous Functions, implies:

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

From Logarithm Tends to Infinity:

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

The result follows from the Push Theorem.


Also see