October 26th, 2016, 03:23 PM  #11  
Senior Member Joined: Jul 2016 From: USA Posts: 108 Thanks: 13  Quote:
And why would it be undefined? 1*1*1*1... It's just repeating itself. It doesn't matter how many times I state it.  
October 26th, 2016, 03:49 PM  #12  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,230 Thanks: 908 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan  
October 26th, 2016, 04:21 PM  #13  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra  Quote:
$$\lim _{n \to \infty} \left(1+\frac{1}{n^2}\right)^n=1$$ because the base heads to 1 "more quickly" than the exponent grows. $$\lim _{n \to \infty} \left(1+\frac{1}{n}\right)^{n^2}=+\infty$$ because the exponent grows "more quickly" than the base heads to 1. $$\lim _{n \to \infty} \left(1+\frac{1}{n}\right)^{n}=e$$ because the base and the exponent remain in balance causing the expression to neither expand nor contract. Each of the indeterminate forms has this conflict between two terms pulling in opposite directions. Like tossing a coin, most of the time one wins over the other: heads or tails. But just occasionally the coin balances on it's edge and the system remains in a sort of unstable equilibrium. Last edited by v8archie; October 26th, 2016 at 04:48 PM.  
October 26th, 2016, 04:31 PM  #14 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,230 Thanks: 908 Math Focus: Wibbly wobbly timeywimey stuff.  
October 27th, 2016, 07:38 AM  #15 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 
a$\displaystyle _{\infty}$=1 if a$\displaystyle _{n}$= n/(1+n) but a$\displaystyle _{n}$ never equals 1. This isn't a paradox. a$\displaystyle _{\infty}$ is a limit, not a$\displaystyle _{\infty}$ evaluated at n=$\displaystyle \infty$. Last edited by zylo; October 27th, 2016 at 07:52 AM. 
October 27th, 2016, 09:23 AM  #16 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra 
$a_\infty$ isn't standard notation. It is misleading too. The concept of evaluating $a_n$ at $n=\infty$ makes no sense because $\infty$ is not a number.

October 27th, 2016, 05:43 PM  #17 
Member Joined: Oct 2016 From: Melbourne Posts: 77 Thanks: 35 
I would start with the function $\displaystyle \begin{align*} f(x) = a^x \!\end{align*}$, then $\displaystyle \begin{align*} f'(x) &= \lim_{h \to 0} \frac{a^{x + h}  a^x}{h} \\ &= \lim_{h \to 0} \frac{a^x\,a^h  a^x}{h} \\ &= a^x\lim_{h \to 0} \frac{a^h  1}{h} \end{align*}$ Now if we want $\displaystyle \begin{align*} f(x) = f'(x) \end{align*}$, in other words, for $\displaystyle \begin{align*} a = \mathrm{e} \end{align*}$ that would require $\displaystyle \begin{align*} \lim_{h \to 0} \frac{a^h  1}{h} &= 1 \\ \lim_{h \to 0} \left( a^h  1 \right) &= \lim_{h \to 0} h \\ \lim_{h \to 0} a^h &= \lim_{h \to 0} \left( 1 + h \right) \\ a &= \lim_{h \to 0} \left( 1 + h \right) ^{\frac{1}{h}} \end{align*}$ and if we define $\displaystyle \begin{align*} n = \frac{1}{h} \end{align*}$ and note that as $\displaystyle \begin{align*} h \to 0 , \, n \to \infty \end{align*}$, this limit is equivalent to $\displaystyle \begin{align*} \mathrm{e} = a = \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right) ^n \end{align*}$ Last edited by skipjack; October 28th, 2016 at 05:43 AM. 
October 28th, 2016, 06:05 AM  #18  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125  Quote:
 
October 28th, 2016, 07:30 AM  #19 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra 
What are you trying to say? Or are you just partially quoting random results from the theory of limits?

October 28th, 2016, 11:29 AM  #20 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra 
Ah! I see what you are trying to say... you think the derivation is incorrect. $$\begin{align*} \lim_{h \to 0} \frac{a^h  1}{h} &= 1 \\ \left(\lim_{h \to 0} h\right) \left(\lim_{h \to 0} \frac{a^h  1}{h}\right) = \lim_{h \to 0} \left(\cancel{h} \frac{a^h  1}{\cancel{h}}\right) &= \lim_{h \to 0} h \\ \lim_{h \to 0} \left(a^h  1\right) &= \lim_{h \to 0} h \\ \lim_{h \to 0} \left(1\right) + \lim_{h \to 0} \left(a^h  1\right) &= \lim_{h \to 0} \left(1\right) + \lim_{h \to 0} \left(h\right) \\ \lim_{h \to 0} \left(1 + a^h  1\right) = \lim_{h \to 0} a^h &= \lim_{h \to 0} \left(1+h\right) \\ \left(\lim_{h \to 0} a^h\right)^{\frac1h} &= \left(\lim_{h \to 0} 1+h\right)^{\frac1h} \\ \lim_{h \to 0} \left(a^h\right)^{\frac1h} = \lim_{h \to 0} a = a &= \lim_{h \to 0} \left(1+h\right)^{\frac1h} \\ \end{align*}$$ 

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