How can I prove that $x^k = O(k^x)$?

Question

Suppose that k is a constant.

I figured out that $k^x \geq x^k$ is true for $x \geq k$. But I couldn't find a way to prove it.

Answer 1

You need to prove that you can find some $C > 0$ such that

$$ x^k \leq C k^x $$

Notice that $\lim_{x \to \infty} \dfrac{ k^x }{x^k } = \infty$ by Lhopitals ( k times) rule and so we can find some $C$ such that $\dfrac{k^x}{x^k} > C $ and thus $Cx^k < k^x $ which means that $x^k = O( k^x ) $

Answer 2

\begin{align} \lim_{x \rightarrow \infty} \frac{x^k}{k^x} = \frac{\infty}{\infty} \end{align} Now, take the derivative of numerator and denominator to obtain $$\lim_{x \rightarrow \infty} \frac{k x^{k-1}}{k^x \log(k)} $$ Repeating the steps above $k$ times (L'Hopital rule) we conclude that the limit equals 0. This means that $k^x$ grows faster than $x^k$. Hence, $x^k \in O(k^x)$.