Limit at Infinity of x^n

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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.

$\blacksquare$


Also see