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Tricky limit. taylor series?

by Felafel
Tags: limit, series, taylor, tricky
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Felafel
#1
Dec22-12, 11:15 AM
P: 171
1. The problem statement, all variables and given/known data

compute the following limit:

## \displaystyle{\lim_{x\to +\infty} x \left((1+\frac{1}{x})^{x} - e \right)} ##

3. The attempt at a solution

i wanted to use the taylor expansion, but didn't know what ##x_0## would be correct, as the x goes to ## \infty##.

also, i tried to use de l'hopital's theorem but it wouldn't work.
how can i do that?
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Sdtootle
#2
Dec22-12, 12:36 PM
P: 17
Is it supposed to be e or e^x?
Hurkyl
#3
Dec22-12, 12:38 PM
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Surely e; this limit would quantify, to first order, how rapidly [itex](1 + 1/x)^x[/itex] converges to its limit e.

vela
#4
Dec22-12, 02:37 PM
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Tricky limit. taylor series?

Try rewriting the limit in terms of z=1/x.
pasmith
#5
Dec22-12, 03:51 PM
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[tex]x\left(\left(1 + \frac1x\right)^x - e\right) = x\left(\exp\left(x\ln \left(1 + \frac1x\right)\right) - e\right)[/tex]

At this point, one can proceed to substitute the Maclaurin series for [itex]\ln(1 + t)[/itex] with [itex]t = x^{-1}[/itex] (we want the limit as [itex]x \to \infty[/itex], so we can assume [itex]0 < t = x^{-1} < 1[/itex]) to get
[tex]x\ln\left(1 + \frac1x\right) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)x^n}[/tex]
and then substitute that into the series for [itex]\exp(x)[/itex] to get
[tex]\exp\left(x\ln\left(1 + \frac1x\right)\right) =
\sum_{k=0}^{\infty} \frac1{k!} \left(\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)x^n}\right)^k = \sum_{m=0}^{\infty} \frac{a_m}{x^m}[/tex]
(everything here is absolutely convergent, so I can add terms in whatever order I want) so that
[tex]x\left(\left(1 + \frac1x\right)^x - e\right) = \left(\sum_{m=0}^{\infty} \frac{a_m}{x^{m-1}} - ex\right) = x(a_0 - e) + a_1 + \sum_{m=2}^{\infty} \frac{a_m}{x^{m-1}}[/tex]
It's clear that the series on the right tends to 0 as [itex]x \to \infty[/itex], so all you need to work out to determine the limit (if any) is [itex]a_0[/itex] and [itex]a_1[/itex]. (You'll need to consider, for each [itex]k[/itex], what the coefficients of [itex]x^0[/itex] and [itex]x^{-1}[/itex] are and add them all together. Actually you can truncate the inner series after [itex]n = 1[/itex], because including [itex]n \geq 2[/itex] doesn't give you any more terms of order [itex]x^0[/itex] and [itex]x^{-1}[/itex] then you already have.

But that is a brute-force method and I'm sure there's a more elegant solution.
SammyS
#6
Dec22-12, 03:52 PM
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Quote Quote by Felafel View Post
1. The problem statement, all variables and given/known data

compute the following limit:

## \displaystyle{\lim_{x\to +\infty} x \left((1+\frac{1}{x})^{x} - e \right)} ##

3. The attempt at a solution

i wanted to use the taylor expansion, but didn't know what ##x_0## would be correct, as the x goes to ## \infty##.

also, i tried to use de l'hopital's theorem but it wouldn't work.
how can i do that?
L'H˘pital's rule works just fine.

[itex]\displaystyle \frac{d}{dx}\left(f(x)\right)^x=\frac{d}{dx}e^{x \ln(f(x))}[/itex]
[itex]\displaystyle =\left(\ln(f(x))+\frac{x}{f(x)}\right)e^{x \ln(f(x))}[/itex]

[itex]\displaystyle =\left(\ln(f(x))+\frac{xf'(x)}{f(x)} \right)\left(f(x)\right)^x[/itex]
Fixed in Edit.
vela
#7
Dec22-12, 04:01 PM
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Quote Quote by SammyS View Post
L'H˘pital's rule works just fine.

[itex]\displaystyle \frac{d}{dx}\left(f(x)\right)^x=\frac{d}{dx}e^{x \ln(f(x))}[/itex]
[itex]\displaystyle =\left(\ln(f(x))+\frac{x}{f(x)}\right)e^{x \ln(f(x))}[/itex]

[itex]\displaystyle =\left(\ln(f(x))+\frac{x}{f(x)} \right)\left(f(x)\right)^x[/itex]
You forgot to apply the chain rule when differentiating ln(f(x)).
SammyS
#8
Dec22-12, 04:55 PM
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Quote Quote by vela View Post
you forgot to apply the chain rule when differentiating ln(f(x)).
DUH !

Thank you vela !
Felafel
#9
Dec23-12, 03:30 AM
P: 171
what if i use the taylor series for ##x_0=1##?
it should become:
##\displaystyle \lim_{x \to \infty}\ x [(2+(x-1)log(4)-1+o((x-1)^2)-e] = \infty * \infty = \infty ##

is it correct?
vela
#10
Dec23-12, 03:37 AM
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Nope
Felafel
#11
Dec23-12, 04:01 AM
P: 171
..maybe it was just de l'hospital then, does this look ok to you?

##\displaystyle \lim{x \to \infty}\ \frac{1}{\frac{1}{x}} ((1+ \frac{1}{x})^x-e) = \frac{0}{0}##
using de l'hopital
##\displaystyle \lim{x \to \infty}\ \frac{1}{\frac{-1}{x^2}} (x(1+\frac{1}{x})^{x-1}*\frac{-1}{x^2}##
which is
## = \infty * \frac{e}{1+0} = \infty ##
vela
#12
Dec23-12, 04:25 AM
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You didn't differentiate correctly. See post #6. You should get a finite answer.
Felafel
#13
Dec23-12, 08:08 AM
P: 171
i can't solve it with derivatives. i eep using de l'hopital but it gets worse and worse.
i thought i could split it into:
##\displaystyle \lim x \to \infty\ x(1+1/x)^x - \displaystyle \lim x \to \infty\ xe ##
and use taylor series for x=0 of the second part, which becomes:
##x*\displaystyle\sum\limits_{k=0}^n \frac{1}{k!} ##
however, i can't figure out the series of the first part
vela
#14
Dec23-12, 12:32 PM
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You don't want to use a Taylor series about x=0. You want to be able to neglect high-order terms which you can't because the limit is for ##x \to \infty##.

Why don't you try my suggestion of rewriting the limit in terms of z=1/x? It'll make applying L'Hopital's more straightforward, and because the limit will be for ##z \to 0##, you can use a Taylor series about ##z=0## if necessary.
Felafel
#15
Dec30-12, 03:01 PM
P: 171
okay, I think I've solved it:
##\lim_{y \to 0}\frac{(1+y)^{\frac{1}{y}}-e}{y}=\lim_{y \to 0} \frac{e^{\frac{1}{y}log(1+y)}-e}{y}##
I find the Maclaurin extension for log(1+y), stopping at the third power

##log(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}##

##\displaystyle{\lim_{y \to 0} \frac{e^{1-\frac{y}{2}-\frac{y^2}{3}}-e }{y}}(=\frac{0}{0})##

## \stackrel{\text{H}}{=} \displaystyle{ \lim_{y \to 0} (-\frac{1}{2}+2\frac{y}{3})e^{1-\frac{y}{2}+\frac{y^2}{2}}}##

##=-\frac{e}{2}##

should be ok now
Dick
#16
Dec30-12, 03:33 PM
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Yes, that's ok now. There's a couple of what are obviously just typos. But nothing important.


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