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chrisplusian
Group Title
find the limit of (n^p)/(e^n) as n approaches infinity
 one year ago
 one year ago
chrisplusian Group Title
find the limit of (n^p)/(e^n) as n approaches infinity
 one year ago
 one year ago

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ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
What is \(p\)?
 one year ago

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

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

ParthKohli Group TitleBest 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.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Am I right? O_O
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
Definitely not the reasoning I'd pick...
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Knew it.
 one year ago

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

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

chrisplusian Group TitleBest ResponseYou've already chosen the best response.0
See 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.
 one year ago

chrisplusian Group TitleBest ResponseYou've already chosen the best response.0
Sorry (1/e^n)
 one year ago

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

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Approaching infinity, not really infinity. I tried to use l'hopital: didn't work. :
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
The almighty Squeeze~
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.2
Isnt the answer 0?
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
ARMAGERD. How could I forget the squeeze
 one year ago

chrisplusian Group TitleBest ResponseYou've already chosen the best response.0
Ok I tried lopitals and it did not work nor did logarithmic properties
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
What's the inequality BTW?
 one year ago

chrisplusian Group TitleBest ResponseYou've already chosen the best response.0
@DLS the answer is zero
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
@DLS how?
 one year ago

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

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

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

chrisplusian Group TitleBest ResponseYou've already chosen the best response.0
@experimentX why would that help?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
L'hopital works use it least floor(p + 1)
 one year ago

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

DLS Group TitleBest ResponseYou've already chosen the best response.2
Well 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: \[y=5x^4\] \[\frac{dy}{dx}=20x^3=>60x^2=>120x=>120\]
 one year ago

chrisplusian Group TitleBest ResponseYou've already chosen the best response.0
Don't understand what you mean @experimentX
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
@chrisplusian dw:1369325101292:dw
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
To 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.
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
And then squeezing :)
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
@terenzreignz yes ceil
 one year ago

DLS Group TitleBest ResponseYou've already chosen the best response.2
@chrisplusian my reply should probably clear it off. @experimentX does it seem right?
 one year ago

chrisplusian Group TitleBest ResponseYou've already chosen the best response.0
the solutions guide says =0 when p>0, and bless than or equal to 2
 one year ago
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