In other words, how to prove:
For all real constants $a$ and $b$ such that $a > 1$,
$$\lim_{n\rightarrow\infty}\frac{n^b}{a^n} = 0$$
I know the definition of limit but I feel that it's not enough to prove this theorem.
Answer
We could prove this by induction on integers $k$:
$$
\lim_{n \to \infty} \frac{n^k}{a^n} = 0.
$$
The case $k = 0$ is straightforward. I will leave the induction step to you. To see how this implies the statement for all real $b$, just note that every real number is less than some integer. In particular, $b \leq \lceil b \rceil$. Thus,
$$
0 \leq \lim_{n \to \infty} \frac{n^b}{a^n} \leq \lim_{n \to \infty} \frac{n^{\lceil b \rceil}}{a^n} = 0.
$$
The first inequality follows since all the terms are positive. The last equality follows from the induction we established previously.
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