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:Limit approaches e Rate Topic: -----

#1 24 December 2011 - 06:55 PM ewmon 


Baryon
I just noticed this. Any reason why

(x+2)x+2/(x+1)x+1 – (x+1)x+1/xx → e

as x → ∞? Does it actually do this?

I checked it in Excel but can't get past x = 141.
If only there were evil people somewhere, insidiously committing evil deeds,
and it were necessary only to separate them from the rest of us and destroy them;
however, the line between good and evil runs through every human heart.
— Aleksandr Solzhenitsyn

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#2 26 December 2011 - 03:23 PM Shadow 


Atom
\frac{(x+2)^{x+2}}{(x+1)^{x+1}} - \frac{(x+1)^{x+1}}{x^x} = \frac{(x+2)^{x+1} \cdot (x+2)}{(x+1)^{x+1}} - \frac{(x+1)^x \cdot (x+1)}{x^x} = (1 + \frac{1}{x+1})^{x+1}\cdot(x+2) - (1 + \frac{1}{x})^x\cdot (x+1) = (1 + \frac{1}{x+1})^{x+1}\cdot(x+1) - (1 + \frac{1}{x})^x\cdot (x+1) + (1 + \frac{1}{x+1})^{x+1}

At this point, I would argue that (1 + \frac{1}{x+1})^{x+1} \sim (1 + \frac{1}{x})^x as x \to \infty \wedge (x+1) = (x+1) \Rightarrow (1 + \frac{1}{x+1})^{x+1} \cdot (x+1) \sim (1 + \frac{1}{x})^x \cdot (x+1) as x \to \infty \Rightarrow \lim_{x\to\infty} (1 + \frac{1}{x+1})^{x+1}\cdot(x+1) - (1 + \frac{1}{x})^x\cdot (x+1) + (1 + \frac{1}{x+1})^{x+1} = \lim_{x\to\infty} (1 + \frac{1}{x+1})^{x+1} = e

But I'd be interested to know if this limit can be solved without the use of asymptotic equalities; it seems L'Hôpital wouldn't lead anywhere.

This post has been edited by Shadow: 26 December 2011 - 03:26 PM

This applies to the above post and any new ideas or media presented in it.

- "Cryptographically secure linear feedback shift register based stream ciphers" -- a phrase that'll get any party started.
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#3 26 December 2011 - 04:12 PM User is online  mississippichem 


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fluorescent protein

View PostShadow, on 26 December 2011 - 03:23 PM, said:

\frac{(x+2)^{x+2}}{(x+1)^{x+1}} - \frac{(x+1)^{x+1}}{x^x} = \frac{(x+2)^{x+1} \cdot (x+2)}{(x+1)^{x+1}} - \frac{(x+1)^x \cdot (x+1)}{x^x} = (1 + \frac{1}{x+1})^{x+1}\cdot(x+2) - (1 + \frac{1}{x})^x\cdot (x+1) = (1 + \frac{1}{x+1})^{x+1}\cdot(x+1) - (1 + \frac{1}{x})^x\cdot (x+1) + (1 + \frac{1}{x+1})^{x+1}

At this point, I would argue that (1 + \frac{1}{x+1})^{x+1} \sim (1 + \frac{1}{x})^x as x \to \infty \wedge (x+1) = (x+1) \Rightarrow (1 + \frac{1}{x+1})^{x+1} \cdot (x+1) \sim (1 + \frac{1}{x})^x \cdot (x+1) as x \to \infty \Rightarrow \lim_{x\to\infty} (1 + \frac{1}{x+1})^{x+1}\cdot(x+1) - (1 + \frac{1}{x})^x\cdot (x+1) + (1 + \frac{1}{x+1})^{x+1} = \lim_{x\to\infty} (1 + \frac{1}{x+1})^{x+1} = e

But I'd be interested to know if this limit can be solved without the use of asymptotic equalities; it seems L'Hτpital wouldn't lead anywhere.


Yeah, I tried L' Hopital for several rounds and kept getting indeterminant forms.
You've come a long way. Remember back when we defined what a velocity meant? Now we are talking about an antisymmetric tensor of second rank in four dimensions.

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#4 30 December 2011 - 11:26 AM ewmon 


Baryon
Shadow and mississippichem, thank you for your derivations and comments. Ashamedly, I was more practiced in math than I am now. Do you think this would be appropriate for mathoverflow.net?
If only there were evil people somewhere, insidiously committing evil deeds,
and it were necessary only to separate them from the rest of us and destroy them;
however, the line between good and evil runs through every human heart.
— Aleksandr Solzhenitsyn

Consider clicking on + if I made you think, or on – if I made you wince ————————————————————————————————►
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