End Behavior of Factorial Function vs Power Function

Question

I'm learning about sequences and I was wondering how to evaluate: $$\lim_{x \to \infty} \frac{10^x}{x!}$$ This gave me an overflow error at sufficiently large numbers on my calculator. In general, is the end behavior of factorials always greater power functions?

Answer 1

Yes. Notice that once $x>10$, we have

$$x!=10!\times11\times12\times\dots\times x>10!\times11^{x-10}$$

Thus,

$$\frac{10^x}{x!}<\frac{10^x}{10!\times11^{x-10}}\stackrel{x\to\infty}{\longrightarrow}0$$

And this is easily extendable to any exponential function.

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As a side note, if

$$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|<1$$

then

$$\lim_{n\to\infty}a_n=0$$

For us, this means that since

$$\lim_{n\to\infty}\left|\frac{\frac{10^{n+1}}{(n+1)!}}{\frac{10^n}{n!}}\right|=\lim_{n\to\infty}\frac{10}{n+1}=0<1$$

then

$$\lim_{n\to\infty}\frac{10^n}{n!}=0$$

Very useful technique for factorial and exponential limits.

Answer 2

If you want to do it in a formal way, you can consider the sequence $a_n=\frac{10^n}{n!}$ and apply the Ratio Test.

It says that if $$\lim_{n\to\infty} \left|{\frac{a_{n+1}}{a_n}}\right| < 1$$ then the limit is $0$.

You see that:

$$\lim_{n\to\infty} \frac{10^{n+1}}{(n+1)!}\frac{n!}{10^n}$$ $$=\lim_{n\to\infty} \frac{10}{n+1} = 0$$

So the limit of the function is $0$.

Answer 3

An easy lower bound for $n!$ is $(n/2)^{n/2}$. This is gotten by looking at the $n/2$ terms in $n!$ which are at least $n/2$.

Therefore, for any $a > 0$, $\frac{a^n}{n!} \lt \frac{a^n}{(n/2)^{n/2}} = (\frac{a^2}{(n/2)})^{n/2} = (\frac{2a^2}{n})^{n/2} $.

By choosing $n$ large enough compared with $2a^2$, say $n > 2ma^2$, we get $\frac{a^n}{n!} \lt (\frac{1}{m})^{n/2} $ and this goes to zero very nicely.

Answer 4

If you want to evaluate $$y=\frac{10^x}{x!}$$ avoiding as much as possible the problem of overflows : take logarithms $$\log(y)=x \log(10)-\log(x!)$$ and use Stirling approximation $$\log(x!)=x (\log (x)-1)+\frac{1}{2} \left(\log (2 \pi )+\log \left({x}\right)\right)+\frac{1}{12 x}+O\left(\frac{1}{x^{5/2}}\right)$$ So, compute first $\log(y)$ and then $y=e^{\log(y)}$.

Let us use $x=100$; this will give $$\log(y)=-133.481\implies y=1.07151\times 10^{-58}$$