big-O proof with power functions

Question

I was wondering if anyone could show a proof for why $a^x$ is $\mathcal{O}(b^x)$ if $a$ and $b$ are constants and $a < b$. In other words, with power functions, does the function with the largest base always eventually overtake a function with a smaller base?

Answer 1

I assume $0<a<b$. Then $$ \frac{a^x}{b^x}=\Bigl(\frac{a}{b}\Bigr)^x\le1\quad\forall x\ge0 $$ since $0<a/b<1$. In fact, $$ \lim_{x\to+\infty}\frac{a^x}{b^x}=0. $$

By the way, $a^x$ is usually called an exponential function, not a power function. This term is used for $x^a$.