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Tricky limit. taylor series? 
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#1
Dec2212, 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? 


#2
Dec2212, 12:36 PM

P: 17

Is it supposed to be e or e^x?



#3
Dec2212, 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.



#4
Dec2212, 02:37 PM

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Tricky limit. taylor series?
Try rewriting the limit in terms of z=1/x.



#5
Dec2212, 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^{m1}}  ex\right) = x(a_0  e) + a_1 + \sum_{m=2}^{\infty} \frac{a_m}{x^{m1}}[/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 bruteforce method and I'm sure there's a more elegant solution. 


#6
Dec2212, 03:52 PM

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[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]Fixed in Edit. 


#7
Dec2212, 04:01 PM

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#8
Dec2212, 04:55 PM

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Thank you vela ! 


#9
Dec2312, 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+(x1)log(4)1+o((x1)^2)e] = \infty * \infty = \infty ## is it correct? 


#10
Dec2312, 03:37 AM

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Nope



#11
Dec2312, 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})^xe) = \frac{0}{0}## using de l'hopital ##\displaystyle \lim{x \to \infty}\ \frac{1}{\frac{1}{x^2}} (x(1+\frac{1}{x})^{x1}*\frac{1}{x^2}## which is ## = \infty * \frac{e}{1+0} = \infty ## 


#12
Dec2312, 04:25 AM

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You didn't differentiate correctly. See post #6. You should get a finite answer.



#13
Dec2312, 08:08 AM

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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 


#14
Dec2312, 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 highorder 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. 


#15
Dec3012, 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 


#16
Dec3012, 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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