:Limit approaches e

[ewmon]

I just noticed this. Any reason why

 

(x+2)^x+2/(x+1)^x+1 – (x+1)^x+1/x^x → e

 

as x → ∞? Does it actually do this?

 

I checked it in Excel but can't get past x = 141.

[Shadow]

[math]\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)[/math] = [math](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}[/math]

 

At this point, I would argue that [math](1 + \frac{1}{x+1})^{x+1} \sim (1 + \frac{1}{x})^x[/math] as [math]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)[/math] as [math] 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[/math]

 

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.

Edited December 26, 2011 by Shadow

[mississippichem]

[math]\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)[/math] = [math](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}[/math]

 

At this point, I would argue that [math](1 + \frac{1}{x+1})^{x+1} \sim (1 + \frac{1}{x})^x[/math] as [math]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)[/math] as [math] 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[/math]

 

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.

 

Yeah, I tried L' Hopital for several rounds and kept getting indeterminant forms.

[ewmon]

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?