Question

find the limit of (n^p)/(e^n) as n approaches infinity

Answer 1

What is \(p\)?

Answer 2

P>0 I really dont even understand why they give that information accept that it shows n remains in the numerator

Answer 3

I tried to use logarithmic properties to find an answer but it just kept running me in circles

Answer 4

\[n^p \times \frac{1}{e^{n}}\]\(e\) 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.

Answer 5

Am I right? O_O

Answer 6

Definitely not the reasoning I'd pick...

Answer 7

Knew it.

Answer 8

correct answer but wrong logic ... think about the opposite e^n/n^p and n->infinity

Answer 9

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

Answer 10

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.

Answer 11

0

Answer 12

Sorry (1/e^n)

Answer 13

Relax... you can either use L'Hôpital
or one of my favourite theorems~ pick one

Answer 14

Approaching infinity, not really infinity.

I tried to use l'hopital: didn't work. :-|

Answer 15

The almighty Squeeze~

Answer 16

Isnt the answer 0?

Answer 17

ARMAGERD. How could I forget the squeeze

Answer 18

Ok I tried lopitals and it did not work nor did logarithmic properties

Answer 19

What's the inequality BTW?

Answer 20

@DLS the answer is zero

Answer 21

Yes,its easy.

Answer 22

Use LH

Answer 23

@DLS how?

Answer 24

My favouraite technique is \[ a = e^{\log a} \\ n^{p} = e^{p \log n } \]

Answer 25

I realise that L'Hôpital may not work after all... at least, not yet...

Use this...
Let \(\large q = \lceil p\rceil\)
such that...
\[\Large 0 \le\frac{n^p}{e^n}\le \frac{n^q}{e^n}\]

Answer 26

q is guaranteed to be a positive integer, means it [the numerator] will (eventually) differentiate to zero...

Answer 27

@experimentX why would that help?

Answer 28

L'hopital works use it least floor(p + 1)

Answer 29

Why deviate from the norm, @experimentX ? :D
Just use ceiling :3

Answer 30

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

Answer 31

Don't understand what you mean @experimentX

Answer 32

@chrisplusian