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February 25th, 2016, 07:57 PM   #1
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Trying to prove limit

Is anyone aware of a simple proof of this:
$\displaystyle \lim\limits_{n\to\infty}\left[\dfrac {(n+x)}{n}\right]^n=e^x$
Edit:
Or failing that, is this identity sufficiently well known that it has a name that I can refer to?

Last edited by Yooklid; February 25th, 2016 at 08:09 PM.
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February 25th, 2016, 08:21 PM   #2
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Update:
I'm guessing that the proof involves taking the log of both sides and then applying L'Hopital's rule:
$\displaystyle \lim\limits_{n\to\infty}n \log\dfrac {(n+x)}{n}=x$
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February 25th, 2016, 09:03 PM   #3
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Well, $\lim \limits_{n \to \infty} \left(1 + \frac1n\right)^n = \mathrm e$ is certainly a well known result (it is sometimes quoted as the definition of $\mathrm e$).

My favourite derivation goes as follows:
For some positive real number $x$ we pick a real number $t$ such that $1 \lt t \lt 1+\frac1x$ which implies that
$${1 \over 1+\frac1x} \lt \frac1t \lt 1$$
This is, by definition valid for all $t$ between $1$ and $1 + \frac1x$, so we can integrate each term over that interval.
$$\newcommand{\integral}[1]{\int_1^{1+\frac1x} {#1} \, \mathrm d t} \integral{1 \over 1+\frac1x} \lt \integral{\frac1t} \lt \integral{1}$$
The outer two integrals have integrands independent of $t$ integrated over an interval of length $\frac1x$ so are just $\frac1x$ times the integrand. The middle integral is the definition of $\log \left(1+\frac1x\right)$. So we have
$$\frac1x \cdot {1 \over 1+\frac1x} = {1 \over x+1} \lt \log \left(1+\frac1x\right) \lt \frac1x$$
Multiplying by $x$ we get
$${x \over x+1} \lt \log \left(1+\frac1x\right)^x \lt 1$$
And then, by the squeeze theorem
$$\lim_{x \to \infty} \log \left(1+\frac1x\right)^x = 1$$
Raising $\mathrm e$ to the power of both sides and taking the limit outside (which we can do because $\mathrm e^x$ is continuous), we have
$$\lim_{x \to \infty} \left(1+\frac1x\right)^x = \mathrm e$$
Then writing $y = {x \over a}$ we get
$$\lim_{x \to \infty} \left(1+{a \over x}\right)^x = \lim_{y \to \infty} \left(1+\frac1y\right)^{ay} = \left(\lim_{y \to \infty} \left(1+\frac1y\right)^y\right)^a= \mathrm e^a$$
Thanks from greg1313 and topsquark
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February 26th, 2016, 10:48 AM   #4
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Very nice! Thank you.
Quite a different form of proof than what I expected.
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February 26th, 2016, 11:40 AM   #5
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$\displaystyle \lim_{n\to\infty}\left[\left(\dfrac{n+x}{n}\right)^n\right]= \lim_{n\to\infty}\left[\left(1+\dfrac{x}{n}\right)^n\right]$

$\displaystyle u=\dfrac{x}{n}$

$\displaystyle \lim_{u\to0}\left[(1+u)^{x/u}\right]=\left(\lim_{u\to0}\left[(1+u)^{1/u}\right]\right)^x=e^x$

Edit: v8archie's got it.
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February 26th, 2016, 11:44 AM   #6
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Alternatively one can expand $n\log\left(1+{x \over n}\right)$ in a Maclaurin series.

Last edited by skipjack; February 26th, 2016 at 03:24 PM.
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February 26th, 2016, 01:36 PM   #7
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Quote:
Originally Posted by Yooklid View Post
Is anyone aware of a > > simple proof < < of this:

"Simple" is a subjective word. I'm still waiting to see a proof of what I imagine would

be the spirit of that type of proof in this thread.


You mentioned L'Hôpital's method in another post.



Let y = lim (n -> oo) [(n + x)/n]^n


y = lim (n -> oo) (1 + x/n)^n


ln(y) = ln[lim (n -> oo) (1 + x/n)^n]


ln(y) = lim (n -> oo) ln(1 + x/n)^n


ln(y) = lim (n -> oo) n*ln(1 + x/n)


ln(y) = lim (n -> oo) [ln(1 + x/n)]/(1/n)


ln(y) = [-x/(n^2)]/[-1/(n^2)]


ln(y) = x


e^[ln(y)] = e^x


y = e^x

Last edited by skipjack; February 26th, 2016 at 03:25 PM.
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February 26th, 2016, 02:45 PM   #8
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Your derivative of $\ln\left(1+{x \over n}\right)$ is incorrect. There is a term missing. You also dropped your limit expression too early.

Since the Maclaurin series solution goes $$\begin{aligned}y &= \lim_{n \to \infty} \left(1+{x \over n}\right)^n \\ \log y &= \lim_{n \to \infty} n \log \left(1+{x \over n}\right) \\ &= \lim_{n \to \infty} n \left( {x \over n} - \frac12\left({x \over n}\right)^2 + \ldots \right) \\ &= \lim_{n \to \infty} \left( x - {x^2 \over 2n} + \ldots \right) \\ &= x \\ y &= \mathrm e^x\end{aligned}$$

I would say it's rather simple.

Last edited by v8archie; February 26th, 2016 at 03:13 PM.
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February 26th, 2016, 04:33 PM   #9
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There's no need for Maclaurin.

$\displaystyle \begin{align*}\ln y &= \ln\lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \\
&= \lim_{h \to 0} \frac{\ln(1 + xh)}{h} \\
&= \frac{d}{dt}\ln(1 + xt)\bigg|_{t=0} \\
&= \frac{x}{1 + xt}\bigg|_{t=0} \\
&= x\end{align*}$
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February 26th, 2016, 04:52 PM   #10
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Quote:
Originally Posted by skipjack View Post
There's no need for Maclaurin.
Clearly not, there are usually many different ways to get a result in maths.
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