Finding $\lim_{x \to \infty} \left[ {x^{x+1} \over (x+1)^x}  - { (x-1)^x\over x^{x-1}}\right]$ 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.
 A: 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,
$$f(x)-f(x-1)= 1/e+O(x^{-1})).$$
A: 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:

A: 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 $x$.)
