TITLE: Finding $\lim_{x \to \infty} \left[ {x^{x+1} \over (x+1)^x}  - { (x-1)^x\over x^{x-1}}\right]$
QUESTION [8 upvotes]: 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.

REPLY [8 votes]: 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,
$$f(x)-f(x-1)= 1/e+O(x^{-1})).$$