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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?

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

Dr. Sonnhard Graubner
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Likai Liu
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  • is it $$\lim_{n\to \infty}\frac{10n!+10^n}{n^{10}+n!}$$? – Dr. Sonnhard Graubner Mar 17 '18 at 09:46
  • @LikaiLiu Note also that l'Hopital is not applicable for sequecences, indeed it is valid for real functions. And also for real functions l'Hopital, in general, is not a good method to solve limits. – user Mar 17 '18 at 09:59
  • @LikaiLiu Please remember that you can choose an answer among the given if the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Mar 19 '18 at 22:54

4 Answers4

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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 $$\frac{10\cdot n!+10^n}{n^{10}+n!}\sim_\infty\frac{10\cdot n!}{n!}=10.$$

Bernard
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  • The lesson is: learn the standard limits to solve also the not standard ones! – user Mar 17 '18 at 10:03
  • I quite agree – one cannot always reinvent the wheel! – Bernard Mar 17 '18 at 10:05
  • Yes of course! I see here that many students are not sufficiently aware about their importance and try to solve all by l'Hopital! – user Mar 17 '18 at 10:07
  • I've also noticed this L'Hospital-mania. In my opinion, insisting on this rule encourages intellectual laziness. – Bernard Mar 17 '18 at 10:11
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if so write your term in the form $$\frac{10+\frac{10^n}{n!}}{\frac{n^{10}}{n!}+1}$$

Dr. Sonnhard Graubner
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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.

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

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

Mohammad Riazi-Kermani
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