ewmon Posted December 24, 2011 Share Posted December 24, 2011 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. Link to comment Share on other sites More sharing options...
Shadow Posted December 26, 2011 Share Posted December 26, 2011 (edited) [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 Link to comment Share on other sites More sharing options...
mississippichem Posted December 26, 2011 Share Posted December 26, 2011 [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. Link to comment Share on other sites More sharing options...
ewmon Posted December 30, 2011 Author Share Posted December 30, 2011 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? Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now