February 25th, 2016, 07:57 PM  #1 
Senior Member Joined: Jul 2008 From: Western Canada Posts: 5,130 Thanks: 46  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. 
February 25th, 2016, 08:21 PM  #2 
Senior Member Joined: Jul 2008 From: Western Canada Posts: 5,130 Thanks: 46 
Update: I'm guessing that the proof involves taking the log of both sides and then applying L'Hopital's rule: $\displaystyle \lim\limits_{n\to\infty}n \log\dfrac {(n+x)}{n}=x$ 
February 25th, 2016, 09:03 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,655 Thanks: 2633 Math Focus: Mainly analysis and algebra 
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 thatThen writing $y = {x \over a}$ we get $$\lim_{x \to \infty} \left(1+{a \over x}\right)^x = \lim_{y \to \infty} \left(1+\frac1y\right)^{ay} = \left(\lim_{y \to \infty} \left(1+\frac1y\right)^y\right)^a= \mathrm e^a$$ 
February 26th, 2016, 10:48 AM  #4 
Senior Member Joined: Jul 2008 From: Western Canada Posts: 5,130 Thanks: 46 
Very nice! Thank you. Quite a different form of proof than what I expected. 
February 26th, 2016, 11:40 AM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,935 Thanks: 1129 Math Focus: Elementary mathematics and beyond 
$\displaystyle \lim_{n\to\infty}\left[\left(\dfrac{n+x}{n}\right)^n\right]= \lim_{n\to\infty}\left[\left(1+\dfrac{x}{n}\right)^n\right]$ $\displaystyle u=\dfrac{x}{n}$ $\displaystyle \lim_{u\to0}\left[(1+u)^{x/u}\right]=\left(\lim_{u\to0}\left[(1+u)^{1/u}\right]\right)^x=e^x$ Edit: v8archie's got it. 
February 26th, 2016, 11:44 AM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,655 Thanks: 2633 Math Focus: Mainly analysis and algebra 
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. 
February 26th, 2016, 01:36 PM  #7 
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  "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'HÃ´pital'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. 
February 26th, 2016, 02:45 PM  #8 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,655 Thanks: 2633 Math Focus: Mainly analysis and algebra 
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. 
February 26th, 2016, 04:33 PM  #9 
Global Moderator Joined: Dec 2006 Posts: 20,627 Thanks: 2077 
There's no need for Maclaurin. $\displaystyle \begin{align*}\ln y &= \ln\lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \\ &= \lim_{h \to 0} \frac{\ln(1 + xh)}{h} \\ &= \frac{d}{dt}\ln(1 + xt)\bigg_{t=0} \\ &= \frac{x}{1 + xt}\bigg_{t=0} \\ &= x\end{align*}$ 
February 26th, 2016, 04:52 PM  #10 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,655 Thanks: 2633 Math Focus: Mainly analysis and algebra  

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