# Tricky limit. taylor series?

1. Dec 22, 2012

### Felafel

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. Dec 22, 2012

### Sdtootle

Is it supposed to be e or e^x?

3. Dec 22, 2012

### Hurkyl

Staff Emeritus
Surely e; this limit would quantify, to first order, how rapidly $(1 + 1/x)^x$ converges to its limit e.

4. Dec 22, 2012

### vela

Staff Emeritus
Try rewriting the limit in terms of z=1/x.

5. Dec 22, 2012

### pasmith

$$x\left(\left(1 + \frac1x\right)^x - e\right) = x\left(\exp\left(x\ln \left(1 + \frac1x\right)\right) - e\right)$$

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

But that is a brute-force method and I'm sure there's a more elegant solution.

6. Dec 22, 2012

### SammyS

Staff Emeritus
L'Hôpital's rule works just fine.

$\displaystyle \frac{d}{dx}\left(f(x)\right)^x=\frac{d}{dx}e^{x \ln(f(x))}$
$\displaystyle =\left(\ln(f(x))+\frac{x}{f(x)}\right)e^{x \ln(f(x))}$

$\displaystyle =\left(\ln(f(x))+\frac{xf'(x)}{f(x)} \right)\left(f(x)\right)^x$

Fixed in Edit.

Last edited: Dec 22, 2012
7. Dec 22, 2012

### vela

Staff Emeritus
You forgot to apply the chain rule when differentiating ln(f(x)).

8. Dec 22, 2012

### SammyS

Staff Emeritus
DUH !

Thank you vela !

9. Dec 23, 2012

### Felafel

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?

10. Dec 23, 2012

### vela

Staff Emeritus
Nope

11. Dec 23, 2012

### Felafel

..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
$= \infty * \frac{e}{1+0} = \infty$

12. Dec 23, 2012

### vela

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

Last edited: Dec 23, 2012
13. Dec 23, 2012

### Felafel

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. Dec 23, 2012

### vela

Staff Emeritus
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.

15. Dec 30, 2012

### Felafel

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. Dec 30, 2012

### Dick

Yes, that's ok now. There's a couple of what are obviously just typos. But nothing important.

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