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 February 25th, 2016, 07:57 PM #1 Senior Member   Joined: Jul 2008 From: Western Canada Posts: 5,194 Thanks: 47 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: 5,194 Thanks: 47 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: 7,681 Thanks: 2659 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 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. $$\newcommand{\integral}{\int_1^{1+\frac1x} {#1} \, \mathrm d t} \integral{1 \over 1+\frac1x} \lt \integral{\frac1t} \lt \integral{1}$$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 February 26th, 2016, 10:48 AM #4 Senior Member   Joined: Jul 2008 From: Western Canada Posts: 5,194 Thanks: 47 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: 7,963 Thanks: 1148 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: 7,681 Thanks: 2659 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
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Quote:
 Originally Posted by Yooklid 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. February 26th, 2016, 02:45 PM #8 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 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: 20,966 Thanks: 2216 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
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Quote:
 Originally Posted by skipjack There's no need for Maclaurin.
Clearly not, there are usually many different ways to get a result in maths. Tags derivation, figure, limit, prove Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post smieci Calculus 6 September 1st, 2014 10:56 AM walter r Real Analysis 4 March 19th, 2014 04:24 PM kaybees17 Real Analysis 2 November 8th, 2013 01:59 AM walter r Real Analysis 8 May 13th, 2013 06:29 AM lesaltersvi Real Analysis 2 February 22nd, 2013 04:47 AM

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