My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum


Thanks Tree4Thanks
Reply
 
LinkBack Thread Tools Display Modes
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.
VisionaryLen is offline  
 
October 25th, 2016, 06:55 PM   #2
Math Team
 
topsquark's Avatar
 
Joined: May 2013
From: The Astral plane

Posts: 2,226
Thanks: 908

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
Originally Posted by VisionaryLen View Post
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.
But $\displaystyle \infty$ is not a number so you can't say that $\displaystyle 1^{\infty}$ is equal to anything. In fact, $\displaystyle 1^{\infty}$ is undefined.

-Dan
topsquark is offline  
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
$$\lim_{y \to 0^+} \left(1 + y\right)^{\frac1y} = e$$
Thanks from topsquark

Last edited by v8archie; October 25th, 2016 at 07:30 PM.
v8archie is offline  
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
$$\lim_{h \to 0} (1+h)^{\frac1h}=e$$
Thanks from topsquark
v8archie is offline  
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.
zylo is offline  
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:
Originally Posted by zylo View Post
[MATH]
$\displaystyle 1^{\infty} = \lim_{n\rightarrow \infty} 1^{n}$ = 1, but that's not the question.
This is false. Specifically, the first expression is not a number and so the first equality is false.

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.
v8archie is offline  
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"!
Country Boy is offline  
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$\displaystyle ^{lnx}$
zylo is offline  
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 non-integer values of $n$ doesn't stop the limit from converging.
v8archie is offline  
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
Quote:
Originally Posted by zylo View Post
L'Hospital's rule is a standard way of determining a limit
The OP didn't as anyone to determine the limit though.
v8archie is offline  
Reply

  My Math Forum > College Math Forum > Calculus

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





Copyright © 2019 My Math Forum. All rights reserved.