chrisplusian
  • chrisplusian
find the limit of (n^p)/(e^n) as n approaches infinity
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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ParthKohli
  • ParthKohli
What is \(p\)?
chrisplusian
  • chrisplusian
P>0 I really dont even understand why they give that information accept that it shows n remains in the numerator
chrisplusian
  • chrisplusian
I tried to use logarithmic properties to find an answer but it just kept running me in circles

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More answers

ParthKohli
  • ParthKohli
\[n^p \times \frac{1}{e^{n}}\]\(e\) 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.
ParthKohli
  • ParthKohli
Am I right? O_O
terenzreignz
  • terenzreignz
Definitely not the reasoning I'd pick...
ParthKohli
  • ParthKohli
Knew it.
experimentX
  • experimentX
correct answer but wrong logic ... think about the opposite e^n/n^p and n->infinity
terenzreignz
  • terenzreignz
I prefer simple cases... \[\Large 2e^n \rightarrow \infty\]\[\Large \frac1{e^n}\rightarrow 0\] \[\Large 2e^n \times \frac1{e^n }\rightarrow 0 \quad \color{red}?\]
chrisplusian
  • chrisplusian
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.
DLS
  • DLS
0
chrisplusian
  • chrisplusian
Sorry (1/e^n)
terenzreignz
  • terenzreignz
Relax... you can either use L'Hôpital or one of my favourite theorems~ pick one
ParthKohli
  • ParthKohli
Approaching infinity, not really infinity. I tried to use l'hopital: didn't work. :-|
terenzreignz
  • terenzreignz
The almighty Squeeze~
DLS
  • DLS
Isnt the answer 0?
ParthKohli
  • ParthKohli
ARMAGERD. How could I forget the squeeze
chrisplusian
  • chrisplusian
Ok I tried lopitals and it did not work nor did logarithmic properties
ParthKohli
  • ParthKohli
What's the inequality BTW?
chrisplusian
  • chrisplusian
@DLS the answer is zero
DLS
  • DLS
Yes,its easy.
DLS
  • DLS
Use LH
ParthKohli
  • ParthKohli
@DLS how?
experimentX
  • experimentX
My favouraite technique is \[ a = e^{\log a} \\ n^{p} = e^{p \log n } \]
terenzreignz
  • terenzreignz
I realise that L'Hôpital may not work after all... at least, not yet... Use this... Let \(\large q = \lceil p\rceil\) such that... \[\Large 0 \le\frac{n^p}{e^n}\le \frac{n^q}{e^n}\]
terenzreignz
  • terenzreignz
q is guaranteed to be a positive integer, means it [the numerator] will (eventually) differentiate to zero...
chrisplusian
  • chrisplusian
@experimentX why would that help?
experimentX
  • experimentX
L'hopital works use it least floor(p + 1)
terenzreignz
  • terenzreignz
Why deviate from the norm, @experimentX ? :D Just use ceiling :3
DLS
  • DLS
Well observe the denominator. Its e^n.No matter how many times you differentiate it,i.e use L-Hopsital,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
  • chrisplusian
Don't understand what you mean @experimentX
experimentX
  • experimentX
@chrisplusian |dw:1369325101292:dw|
terenzreignz
  • terenzreignz
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.
terenzreignz
  • terenzreignz
And then squeezing :)
experimentX
  • experimentX
@terenzreignz yes ceil
DLS
  • DLS
@chrisplusian my reply should probably clear it off. @experimentX does it seem right?
chrisplusian
  • chrisplusian
the solutions guide says =0 when p>0, and bless than or equal to 2

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