In this proof, we first restrict to when $x$ and $a$ are integers and only later lift this restricton.

Let $a > 0$ be an integer, let $b > 1$ be real, and let $x$ be an integer.

Consider the following inequality $$ \left( 1 + {1 \over x} \right)^a \le 1 + {a \over x} \left( 1 + {1 \over x} \right)^{a-1} $$ If $x \ge 2$ then we have $$ \left( 1 + {1 \over x} \right)^a \le 1 + {a \over x} \left( {3 \over 2} \right)^{a-1} . $$ Define $X$ to be the greater of $2$ and $\lceil a (3/2)^{a-1} / (1 - \sqrt{b}) \rceil$ when $x > X$ we have $$ \left( 1 + {1 \over x} \right)^a \le \sqrt{b}. $$

Rewrite $x^a / b^x$ as follows when $x > X$ $$ {x^a \over b^x} = {X^a \over b^X} \prod_{n=X}^x \left( 1 + {1 \over n} \right)^a {1 \over b} $$ By the inequality established above, each term in the product will be bounded by $1 / \sqrt{b}$ hence $$ {x^a \over b^x} \le {X^a \over b^X} {1 \over (\sqrt{b})^{x - X}} $$ Since $b > 1$ it is also the case that $\sqrt{b} > 1$ hence we have the inequality $$ (\sqrt{b})^n \ge 1 + n (\sqrt{b} - 1) $$ Combining the last two inequalities yields the following: $$ {x^a \over b^x} \le {X^a \over b^X} \le {1 \over 1 + (x - X) (\sqrt{b} - 1)} $$ From this, it follows that $\lim_{x \to \infty} x^a / b^x = 0$ when $a$ and $x$ are integers.

Now we lift the restriction that $a$ be an integer. Since the power function is increasing, $x^a / b^x \le x^{\lceil a \rceil} / b^x$ so we have $\lim_{x \to \infty} x^a / b^x = 0$ for real values of $a$ as well.

To lift the restriction on $x$ let us write $x = x_1 + x_2$ where $x_1$ is an integer and $0 \le x_2 < 1$ Then we have $$ {x^a \over b^x} = {x_1^a \over b^{x_1}} \left( {x_1 + x_2 \over x_1} \right)^a b^{-x_2} $$ If $x > 2$ then $(x_1 + x_2) / x_2 < 1.5$ Since $x_2 \ge 0, b^{-x_2} \le 1$ Hence, for all real $x > 2$ we have $$ {x^a \over b^x} \le 1.5^a {x_1^a \over b^{x_1}}$$ From this inequality, it follows that $\lim_{x \to \infty} x^a / b^x = 0$ for real values of $x$ as well.