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 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
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 Originally Posted by VisionaryLen 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 October 25th, 2016, 07:17 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 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. October 25th, 2016, 07:42 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 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 October 26th, 2016, 12:03 AM #5 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 $\displaystyle \lim_{x\rightarrow 0}(1+x)^{1/x} = \lim e^{\ln(1+x)^{1/x}} = e^{\lim \ln(1+x)^{1/x}}$ $\displaystyle \lim \ln(1+x)^{1/x} = \lim \frac{\ln(1+x)}{x}$ = 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
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 Originally Posted by zylo [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. 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: 126 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: 7,690 Thanks: 2669 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. October 26th, 2016, 06:52 AM   #10
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 Originally Posted by zylo L'Hospital's rule is a standard way of determining a limit
The OP didn't as anyone to determine the limit though. Tags approach, x1 or x 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 matthius Calculus 0 February 20th, 2012 07:26 AM JohnA Algebra 2 February 19th, 2012 09:29 AM tiba Math Books 0 February 19th, 2012 07:03 AM cmmcnamara Advanced Statistics 4 February 10th, 2010 05:49 AM

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