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The limit is $\lim_{x \to \infty} \left[ {x^{x+1} \over (x+1)^x} - { (x-1)^x\over x^{x-1}}\right]$

Experimentally, this limit appears to converge to ${1 \over e}$, but I can't figure out how to solve it.

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    You will have to take logarithms at some point and come up with an expression whose limit is -1. But since this is the limit for the logarithm, the limit for the original expression should be $\exp(-1)$.2010-08-14

3 Answers 3

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A fairly mechanical approach is to write the limit as $\lim_{x\to\infty}(f(x)-f(x-1))$ where $f(x)=\frac{x^{x+1}}{(x+1)^x}.$ Then $\log f(x)=(x+1)\log x - x\log(x+1) = \log x - x \log(1+1/x)$ and so $\log f(x)= \log x - 1 + 1/(2x) + O(x^{-2})$ as $x\to\infty$ (using the Maclaurin series for $\log(1+t)$). Therefore $f(x) = (x/e)(1+1/(2x)+O(x^{-2}))=x/e+1/(2e)+O(x^{-1})$ and so $f(x-1) =(x-1)/e+1/(2e) +O((x-1)^{-1}).$ Subtracting, $f(x)-f(x-1)= 1/e+O(x^{-1})).$

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Making the change of variables $\; z = 1/x \;$ and collecting together like exponents yields

$\quad\quad\quad\quad \displaystyle \lim_{z\to 0^+} {\frac{(1+z)^{\large -1/z} - (1-z)^{\large 1/z}}{z}}\; = \;\lim_{z\to 0^+} {\frac{f(-z)-f(z)}{z}} $

Applying the well-known $\rm exp$ and $\rm log$ taylor series to $f(z) = (1-z)^{\large 1/z}\; = \; e^{\large {\rm log}(1-z)/z}$

$f(z) = e^{\large -1-\frac{z}{2}+\;\cdots} = e^{-1} (1 + (-\frac{z}{2} +\;\cdots) + (-\frac{z}{2} +\cdots)^2 + \;\cdots) = \frac{1}{e} - \frac{z}{2e} + O(z^2)$

$\displaystyle {\rm Therefore}\quad \frac{f(-z)-f(z)}{z}\; = \;\frac{1}{e} + O(z),\ \ $ as confirmed by Macsyma below:

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HINT You can write the difference as $x \left( y^{-x} - z^x \right)=x\left(1-(yz)^x\right)/y^z$, where $y$ and $z$ are rational functions of $x$ such that the limiting value of $y^x$ (and $z^x$) is well-known, finite, and non-zero. Now focus on finding the limit of $x\left(1-(yz)^x\right)$.

(Restricting $x$ to integers and invoking the Binomial Theorem helps, but that only shows what the limit --if it exists-- would have to be, leaving open the possibility that the limit might not exist when considering non-integer $x$.)

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    Well, the binomial theorem still works for noninteger exponents (we now have an infinite series because none of the binomial coefficients vanish), though it no longer applies if the variable of interest is both in the base and the exponent.2010-08-14