Intuition behind $\lim _{x \to\infty} \frac{x^n}{a^x}=0$

Question

Could someone intuitively explain why

$$ \lim _{x \to\infty} \frac{x^n}{a^x}=0 \quad(n \in \mathbb{N}, a>1) $$

Don't we get $\frac{\infty}{\infty}$ in this case, which is indeterminant?

Answer 1

Since $a^x = \exp(x\log(a))$ and $a>1$ so that $c=\log(a)>0$, if we set $t=x/c$ then $$ \frac{x^n}{a^x} = \frac{c^nt^n}{\exp(t)} \implies \lim_{x\to \infty}\frac{x^n}{a^x} = c^n\lim_{t \to \infty} \frac{t^n}{\exp(t)}. $$

Since $c^n$ is a constant independent of $t$ and $x$, it therefore suffices to consider the case $a=e$, and $e^x = \exp(x)$.

Now the Taylor series for $\exp(x)$ gives $\exp(x) = \sum_{k=0}^\infty \frac{x^k}{k!}$, and for $x>0$, all of the terms in the infinite sum are positive, and so $\exp(x)$ is greater than any single term. Thus we have, for any $m$, $\exp(x)>x^m/m!$. But then given $n$, we may pick $m=n+1$ say, and we see $$ \frac{x^n}{e^x} <\frac{x^n}{x^{n+1}/(n+1)!} = \frac{(n+1)!}{x} \to 0 \text{ as } t\to \infty. $$

Answer 2

An indeterminate form doesn't mean the limit doesn't exist, it just means that you can't tell straight away what the limit is. Broadly speaking, a $\frac{\infty}{\infty}$ limit can do any of the following:

1. Have the numerator grow faster than the denominator, and tend to infinity;

2. Have the denominator grow faster than the numerator, and tend to zero;

3. Have both grow at a similar pace, and tend to a real limit;

4. Have the two cancel out in ways that causes the limit to not exist.

To understand the $\frac{x^n}{a^x}$ case, imagine we both start with one dollar, and every day you earn one more dollar while I double the money I currently have. As time passes, both of us will have more money and in an infinite amount of time we'll both earn infinite amounts. But if we look at the ratio of how much money we have as the days pass, you'll see that when you have \$$n$, I have \$$2^n$, and so your balance is going to become an ever-decreasing fraction of mine - on day 10 it's \$10 versus \$1024, on day 100 it's \$100 versus \$1,267,650,600,228,229,401,496,703,205,376.

As it turns out, as long as you're earning money polynomially fast while I'm earning it exponentially fast, it doesn't matter what the specific values are - you could be getting \$$n^{1000}$ per day while I'm compounding at \$$(1.0001)^n$, even though it's going to take me a long time to catch up with you eventually the amount I'm gaining is going to be more than you and after that the same thing happens and the ratio between our balances will dwindle to nothing.

And just to demonstrate trivially how indeterminate forms can do various things, consider the following cases:

1. $\lim_{x \rightarrow \infty} \frac{2x}{x^2}$

2. $\lim_{x \rightarrow \infty} \frac{2x^2}{x}$

3. $\lim_{x \rightarrow \infty} \frac{2x}{x}$

All of these are technically of the form $\frac{\infty}{\infty}$, but you can easily cancel out factors of $x$ to see that the first one tends to zero, the second to infinity, and the third is actually a constant 2.