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chrisplusian
 one year 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
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0Definitely not the reasoning I'd pick...

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

chrisplusian
 one year 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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