Visual proof that $\lim_{x\to \infty} \frac{x^r}{a^x} = 0$

Question

I am looking for a visual proof (in this sense: https://mathoverflow.net/questions/8846/proofs-without-words) for the fact that

$$ \lim_{x\to \infty} \frac{x^r}{a^x} = 0 $$

($r > 0,a>1$)

Or at least for a simple (high school level) argument that makes this relation plausible (but it doesn't have to be a rigerous mathematical proof - I know to proof it rigerously, but that's not what I am looking for).

Answer 1

Every time $x$ increases by $1$, the bottom gets multiplied by $a$ and the top gets multiplied by $$\left(\frac{x+1}{x}\right)^r = \left(1+\frac{1}{x}\right)^r \approx 1+\frac{r}{x}$$

So the bottom keeps growing by the same factor while the factor that the top grows by gets smaller and smaller, and will eventually get smaller than $a$. In the long run (i.e. as $x \to\infty$) the bottom will win out, driving the whole expression to $0$. (This does make an appeal to a binomial theorem approximation.)

It's not intuitive, and probably doesn't have the "Aha!" appeal of a visual proof, but I think it would help high schoolers start to think about comparing different rates of growth and help them build their intuition.

And if I were presenting this to high schoolers I would definitely start off showing how it worked in a specific simple example, like $\frac{x^2}{2^x}$. Write ten terms of $x^2$, note that they're increasing, but then write ten terms of $2^x$ and note how the crush the terms on the top. Then make the generalization, and maybe note that if $a$ is close to $1$ and $r$ is large it will take more terms for the bottom to win, but it will eventually, and we are going to $\infty$.

Answer 2

I guess that we can do the following: $$ 0 = \lim_{x \rightarrow \infty} \frac{x^r}{a^x} = \lim_{x \rightarrow \infty} \frac{\ln x^r}{\ln a^x} = \lim_{x \rightarrow \infty} \frac{r \ln x}{x \ln a} = \frac r {\ln a} \lim_{x \rightarrow \infty} \frac{\ln x}{x} = \lim_{x \rightarrow \infty} \frac{\ln x}{x} $$

Using the last limit we can make visual proof.

Answer 3

Let $a^{1/r}=1+b.$ Then $b>0$ if $a>1$ and $r>0.$ For any $y$ let $[y]$ denote the largest integer not exceeding $y.$

Apply a result from the Binomial Theorem: $(1+b)^n>b^2\binom {n}{2}$ when $b>0$ and $2<n\in \mathbb N$.

For $\frac {x}{r}\geq 3$ we have $$0<\frac {x^r }{a^x}=\left(\frac {x}{(1+b)^{x/r}}\right)^r\leq \left(\frac {x}{(1+b)^{[x/r]}}\right)^r<$$ $$<\left(\frac {x}{b^2\cdot \binom {[x/r]}{2}}\right)^r<$$ $$< \left(2rb^{-2}\cdot \frac {\frac {x}{r}}{(\frac{x}{r}-1)(\frac{x}{r}-2)}\right)^r$$