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anonymous
 3 years ago
find the limit of (n^p)/(e^n) as n approaches infinity
anonymous
 3 years ago
find the limit of (n^p)/(e^n) as n approaches infinity

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anonymous
 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

anonymous
 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... \[\Large 2e^n \rightarrow \infty\]\[\Large \frac1{e^n}\rightarrow 0\] \[\Large 2e^n \times \frac1{e^n }\rightarrow 0 \quad \color{red}?\]

anonymous
 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

anonymous
 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?

anonymous
 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...

anonymous
 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: \[y=5x^4\] \[\frac{dy}{dx}=20x^3=>60x^2=>120x=>120\]

anonymous
 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?

anonymous
 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
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