A community for students.
Here's the question you clicked on:
 0 viewing
chrisplusian
 3 years ago
find the limit of (n^p)/(e^n) as n approaches infinity
chrisplusian
 3 years ago
find the limit of (n^p)/(e^n) as n approaches infinity

This Question is Closed

chrisplusian
 3 years ago
Best ResponseYou've already chosen the best response.0P>0 I really dont even understand why they give that information accept that it shows n remains in the numerator

chrisplusian
 3 years ago
Best ResponseYou've already chosen the best response.0I tried to use logarithmic properties to find an answer but it just kept running me in circles

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0\[n^p \times \frac{1}{e^{n}}\]\(e\) is a constant. Oh, and \(n^p \to \infty \) and \(\dfrac{1}{e^n} \to 0\). That can only mean that their product is approaching 0.

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0Definitely not the reasoning I'd pick...

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0correct answer but wrong logic ... think about the opposite e^n/n^p and n>infinity

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0I prefer simple cases... \[\Large 2e^n \rightarrow \infty\]\[\Large \frac1{e^n}\rightarrow 0\] \[\Large 2e^n \times \frac1{e^n }\rightarrow 0 \quad \color{red}?\]

chrisplusian
 3 years ago
Best ResponseYou've already chosen the best response.0See i don't understand becasue if you look at it as (N^p)*(1/e^n) then the n^p is infinity and the !/e^n is zero. That gives you infinity times zero which is an indeterminate form.

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0Relax... you can either use L'Hôpital or one of my favourite theorems~ pick one

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Approaching infinity, not really infinity. I tried to use l'hopital: didn't work. :

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0The almighty Squeeze~

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0ARMAGERD. How could I forget the squeeze

chrisplusian
 3 years ago
Best ResponseYou've already chosen the best response.0Ok I tried lopitals and it did not work nor did logarithmic properties

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0What's the inequality BTW?

chrisplusian
 3 years ago
Best ResponseYou've already chosen the best response.0@DLS the answer is zero

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0My favouraite technique is \[ a = e^{\log a} \\ n^{p} = e^{p \log n } \]

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0I realise that L'Hôpital may not work after all... at least, not yet... Use this... Let \(\large q = \lceil p\rceil\) such that... \[\Large 0 \le\frac{n^p}{e^n}\le \frac{n^q}{e^n}\]

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0q is guaranteed to be a positive integer, means it [the numerator] will (eventually) differentiate to zero...

chrisplusian
 3 years ago
Best ResponseYou've already chosen the best response.0@experimentX why would that help?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0L'hopital works use it least floor(p + 1)

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0Why deviate from the norm, @experimentX ? :D Just use ceiling :3

DLS
 3 years ago
Best ResponseYou've already chosen the best response.2Well observe the denominator. Its e^n.No matter how many times you differentiate it,i.e use LHopsital,the denominator WON'T CHANGE at all.That is the first thing.Now I say,use L hospital infinity times(not really) so if you keep on differentiating..a time will come when numerator turns out to be 0. so 0/1=1. Example: \[y=5x^4\] \[\frac{dy}{dx}=20x^3=>60x^2=>120x=>120\]

chrisplusian
 3 years ago
Best ResponseYou've already chosen the best response.0Don't understand what you mean @experimentX

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0@chrisplusian dw:1369325101292:dw

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0To simplify things, we best make sure the exponent p in the numerator is an (positive, but that's granted) integer. Taking \(\Large q = \lceil p\rceil\) helps with that.

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0And then squeezing :)

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0@terenzreignz yes ceil

DLS
 3 years ago
Best ResponseYou've already chosen the best response.2@chrisplusian my reply should probably clear it off. @experimentX does it seem right?

chrisplusian
 3 years ago
Best ResponseYou've already chosen the best response.0the solutions guide says =0 when p>0, and bless than or equal to 2
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.