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chrisplusianBest 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
 11 months ago

chrisplusianBest 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
 11 months ago

ParthKohliBest 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.
 11 months ago

terenzreignzBest ResponseYou've already chosen the best response.0
Definitely not the reasoning I'd pick...
 11 months ago

experimentXBest ResponseYou've already chosen the best response.0
correct answer but wrong logic ... think about the opposite e^n/n^p and n>infinity
 11 months ago

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

chrisplusianBest 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.
 11 months ago

terenzreignzBest ResponseYou've already chosen the best response.0
Relax... you can either use L'Hôpital or one of my favourite theorems~ pick one
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
Approaching infinity, not really infinity. I tried to use l'hopital: didn't work. :
 11 months ago

terenzreignzBest ResponseYou've already chosen the best response.0
The almighty Squeeze~
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
ARMAGERD. How could I forget the squeeze
 11 months ago

chrisplusianBest ResponseYou've already chosen the best response.0
Ok I tried lopitals and it did not work nor did logarithmic properties
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
What's the inequality BTW?
 11 months ago

chrisplusianBest ResponseYou've already chosen the best response.0
@DLS the answer is zero
 11 months ago

experimentXBest ResponseYou've already chosen the best response.0
My favouraite technique is \[ a = e^{\log a} \\ n^{p} = e^{p \log n } \]
 11 months ago

terenzreignzBest 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... \[\Large 0 \le\frac{n^p}{e^n}\le \frac{n^q}{e^n}\]
 11 months ago

terenzreignzBest 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...
 11 months ago

chrisplusianBest ResponseYou've already chosen the best response.0
@experimentX why would that help?
 11 months ago

experimentXBest ResponseYou've already chosen the best response.0
L'hopital works use it least floor(p + 1)
 11 months ago

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

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

chrisplusianBest ResponseYou've already chosen the best response.0
Don't understand what you mean @experimentX
 11 months ago

experimentXBest ResponseYou've already chosen the best response.0
@chrisplusian dw:1369325101292:dw
 11 months ago

terenzreignzBest 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.
 11 months ago

terenzreignzBest ResponseYou've already chosen the best response.0
And then squeezing :)
 11 months ago

experimentXBest ResponseYou've already chosen the best response.0
@terenzreignz yes ceil
 11 months ago

DLSBest ResponseYou've already chosen the best response.2
@chrisplusian my reply should probably clear it off. @experimentX does it seem right?
 11 months ago

chrisplusianBest ResponseYou've already chosen the best response.0
the solutions guide says =0 when p>0, and bless than or equal to 2
 11 months ago
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