# Thread: Sequence's limit and indeterminate form, kinda need help..

1. ## Sequence's limit and indeterminate form, kinda need help..

Hello, first of all i might excuse myself for my poor english, i'm Italian and my native language is far from being "english-friendly";

down to the question itself i had a test at university where i was called to solve this limit:

i might also want to specify that we could not use De L'Hopital or other "advanced tools" as we're studying "Sequences' limits" and not "Functions' limits", so we have to basically reduce this to a "collection of known limits" to be able to specify the value it assumes when n->+infinity.

Well, i know by heart that the numerator is "bigger" than the denominator so the limit is going to be infinite, but i should not conclude the exercise that way and i just can't find a way to simplify it.
How would you guys go with that?

SkyWolf.

2. Let $u_n:=\frac{n!+n2^n}{n+4^n}$. We have $u_n\geq \frac{n!}{2\cdot 4^n}$ so we only have to show that $\lim_{n\to\infty}\frac{n!}{4^n}=+\infty$. If we denote by $v_n$ this sequence, it's an increasing one. You only have to show that it is not bounded.

3. Originally Posted by girdav
... We have $u_n\geq \frac{n!}{2\cdot 4^n}$ so we only have to show that $\lim_{n\to\infty}\frac{n!}{4^n}=+\infty$. ...
Thanks for the answer, but why did you pick $\frac{n!}{2\cdot 4^n}$? There is any particular reason?
Why $2\cdot 4^n$ as denominator, couldn't we get just $4^n$?

SkyWolf.

4. I wrote $n+4^n\leq 4^n+4^n$.

5. Originally Posted by girdav
I wrote $n+4^n\leq 4^n+4^n$.
Derp I am soooo dumb.

Many thanks,
SkyWolf.