Question

Evaluation of limit as n approaches infinity of (2*n^2)/(2^n)

Answer 1

I need some way to show that the answer to this is 0, I'm just not sure how

Answer 2

apply L'hopital 2 times

Answer 3

nothing much happens to the denominator
but the numerator reduces to a constant

Answer 4

\[\large \lim\limits_{n\to \infty} \dfrac{c}{2^n} = 0\]

Answer 5

Not sure why I didn't see that, but thank you for pointing that out. Now would there be an easy way to do that with basically the same equation:
\[ \frac{ k*n^k }{ 2^n }\]
where k is an arbitrary natural number?

Answer 6

or would it just be a matter of using L'Hopital's rule k number of times?

Answer 7

same thing,
if you wait long enough, the exponential overtakes ANY polynomial

Answer 8

\[\large \lim\limits_{n\to \infty}\dfrac{n^{1000000000000000}}{2^n} = 0\]

Answer 9

and yes apply L'opital a trillion times :)

Answer 10

Okay. thank you very much. Not sure why I didn't realize that, but thank you for the help!

Answer 11

np :) can you think of any other way to prove this without touching L'opital ?

Answer 12

I know there's a way but isn't it a lot more painful then L'Hopital's, or am I thinking something else?

Answer 13

i remember a prof saying about this :
you can avoid applying L'opital trillion times if you are clever enough.. but he didn't show the other method... i also gave up after trying for a while

Answer 14

from his talk, it appears the other way is more intuitive and fast
let me pull up that video

Answer 15

I know there are shortcuts if certain polynomials are the same, but otherwise a lot of the higher level limit work I did wasn't too fun

Answer 16

I got a question about limits, away from your original question, sorry to intrude if I am, but \[\lim_{x \rightarrow {\infty}}~~~ 1^\infty = 1\] I know this is an intermediate form, but why can't we just say it = 1? What was the reason for this?

Answer 17

The same old approaching from left and right?

Answer 18

in the context of my question or just in general?

Answer 19

General I guess

Answer 20

My guess is that there isn't a particular reason, it's just to show that it is still changing and the problem isn't completely finished yet. Just a guess

Answer 21

It's been troubling me, maybe have to take multiple limits?

Answer 22

http://mathforum.org/library/drmath/view/56006.html

Answer 23

the classic example is

\[\large \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e \ne 1\]

even though \(\large \lim\limits_{n\to\infty}(1+\frac{1}{n}) = 1 \) and \(\large \lim\limits_{n\to\infty }n = \infty \)

Answer 24

@goalie2012 watch below video between `44-46` minutes :
http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-19-introduction-to-the-laplace-transform/

he says we can do it by applying L'opital only once but im not able to see how thats possible :/

Answer 25

I don't either unless you have to move a few things around after you use the rule once? or theres a different rule you can use after that? I'm not sure.

Answer 26

yeah its been a mystery for me from the day i took that course haha! hope he was not thinking about taylor series >.<

Answer 27

I agree. Those aren't fun at all... Might have to show that to a couple people to see what they think