October 25th, 2016, 06:30 PM  #1 
Senior Member Joined: Jul 2016 From: USA Posts: 108 Thanks: 13  Why does (1+x)^(1/x) approach e?
Why does (1+x)^(1/x) = e as x approaches 0, as shown in this graph? https://www.desmos.com/calculator/d1bvs8tqlu 1 to the power of anything, including infinity, should be 1. 1*1*1*1*1*...=1 It's just repeating itself. It's like saying 1+0+0+0+0... nothing will ever change. So why shouldn't this equation approach 1 instead of e? I have my own personal theories, but I would like to get input from everyone else as I am sure there will be some pretty good answers from this community. Thanks. 
October 25th, 2016, 06:55 PM  #2  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,226 Thanks: 908 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan  
October 25th, 2016, 07:17 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra 
It's not one to the power of anything though. It's numbers greater than one raised to very large powers. For any $x \gt 0$ we and any $t$ such that $1 \lt t \lt 1+\frac1x$ we have (by taking the reciprocals) $$\frac{x}{x+1} \lt \frac1t \lt 1$$ Since this is valid over the interval $(1, 1+\frac1x)$ we can integrate over that interval to get $$\int_{1}^{1+\frac1x} \frac{x}{x+1} \, \mathrm dt \lt \int_{1}^{1+\frac1x} \frac1t \, \mathrm dt \lt \int_{1}^{1+\frac1x} \mathrm dt$$ Evaluating the integrals gives $$\frac1{1+x} \lt \ln{\left(1 + \frac1x\right)} \lt \frac1x$$ Multiplying by $x$ and using the logarithmic identity $a\ln b = \ln b^a$ we have $$\frac{x}{1+x} \lt \ln{\left(1 + \frac1x\right)^x} \lt 1$$ As $x \to +\infty$ the left and right members both tend to $1$, so by the squeeze theorem $$\lim_{x \to +\infty} \ln{\left(1 + \frac1x\right)^x} = 1$$ and thus $$\lim_{x \to +\infty} \left(1 + \frac1x\right)^x = e$$ Writing $y=\frac1x$ we have $$\lim_{y \to 0^+} \left(1 + y\right)^{\frac1y} = e$$ Last edited by v8archie; October 25th, 2016 at 07:30 PM. 
October 25th, 2016, 07:42 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra 
Alternatively, for $f(x)=\ln(1+x)$ $$f'(0)=\lim_{h \to 0} \frac{\ln(1+h)  \ln(1)}{h}=\frac1{1+0}$$ Thus, by $a\ln b=\ln b^a$ $$\lim_{h \to 0} \ln(1+h)^{\frac1h}=1$$ And so $$\lim_{h \to 0} (1+h)^{\frac1h}=e$$ 
October 26th, 2016, 12:03 AM  #5 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 
$\displaystyle \lim_{x\rightarrow 0}(1+x)^{1/x} = \lim e^{\ln(1+x)^{1/x}} = e^{\lim \ln(1+x)^{1/x}}$ $\displaystyle \lim \ln(1+x)^{1/x} = \lim \frac{\ln(1+x)}{x}$ = 1, by L'Hôpital's rule. $\displaystyle \lim_{x\rightarrow 0}(1+x)^{1/x}$ = e $\displaystyle 1^{\infty} = \lim_{n\rightarrow \infty} 1^{n}$ = 1, but that's not the question. Last edited by skipjack; October 26th, 2016 at 02:06 AM. 
October 26th, 2016, 04:08 AM  #6  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra  Quote:
I also seriously doubt the ability of l'Hôpital's rule to explain in any meaningful way why the limit is as it is. I will accept that the derivative demonstration I gave suffers from the same problem. The first shows that the logarithm is always between $\frac{x}{1+x}$ and $1$, giving the result. Last edited by v8archie; October 26th, 2016 at 04:23 AM.  
October 26th, 2016, 05:52 AM  #7 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
How you prove that depends upon exactly how you have defined "e"!

October 26th, 2016, 06:40 AM  #8 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 
L'Hospital's rule is a standard way of determining a limit such as 0/0 or $\displaystyle \infty/\infty$. $\displaystyle a_{\infty} = \lim_{n\rightarrow \infty}$a$\displaystyle _{n}$ by definition e$\displaystyle ^{x}$ is the inverse of lnx by definition so that x=e$\displaystyle ^{lnx}$ 
October 26th, 2016, 06:49 AM  #9 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra 
All the definitions are equal (obviously) so you can start anywhere. But the question wasn't to prove it, but to explain it. Starting with $e=\lim_{n \to \infty} \left(1+\frac1n\right)^n$ as I suspect you are hinting at is unlikely to explain much. Curiously, the biggest hurdle in a proof with that starting point has is to prove that including noninteger values of $n$ doesn't stop the limit from converging. 
October 26th, 2016, 06:52 AM  #10 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra  

Tags 
approach, x1 or x 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Hessian approach  matthius  Calculus  0  February 20th, 2012 07:26 AM 
How to approach this problem  JohnA  Algebra  2  February 19th, 2012 09:29 AM 
Different Geometry Approach  tiba  Math Books  0  February 19th, 2012 07:03 AM 
How do I approach this?  cmmcnamara  Advanced Statistics  4  February 10th, 2010 05:49 AM 