My Math Forum Trying to prove limit

 Calculus Calculus Math Forum

 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 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
 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

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: 7,655 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
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 Display Modes Linear 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

 Contact - Home - Forums - Cryptocurrency Forum - Top