How can i get the limit of $(10(n!)+10^n)/(n^{10}+n!)$ when $n$ goes to $\infty$?

Question

I am working on this limit question right now. I tried to use something like l'Hôpitals rule but this question has Factorial terms. Is there any trick that i can use in this question?

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

Answer 1

if so write your term in the form $$\frac{10+\frac{10^n}{n!}}{\frac{n^{10}}{n!}+1}$$

Answer 2

Yes. L'Hospital's rule is not the alpha and omega of limits computation! Much more powerful, you can use asymptotic calculus, , in particular equivalents:

We know that, for any $a\in\mathbf R$, $a^n=_\infty o(n!)$, $n^t=_\infty o(n!)$, so that $\;10\cdot n!+10^n\sim_\infty10\cdot n!$, $\;n^{10}+n!\sim_\infty n!$, hence $$\frac{10\cdot n!+10^n}{n^{10}+n!}\sim_\infty\frac{10\cdot n!}{n!}=10.$$

Answer 3

You need to figure out which is the "stronger" term, in this case note that by standard limits

$$\frac{a^n}{n!}\to 0 \qquad \frac{n^a}{n!}\to 0$$

indeed by ratio test

• $\frac{a^{n+1}}{(n+1)!}\frac{n!}{a^n}=\frac{a}{n+1}\to 0$

• $\frac{(n+1)^{a}}{(n+1)!}\frac{n!}{n^a}=\frac{(1+1/n)^a}{n+1}\to 0$

then consider

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

and use the previuos results.

Answer 4

Note that $n!$ is the dominating term as $n\to \infty$

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