TITLE: Is this a new expression for $e$?
QUESTION [8 upvotes]: Let $f(x) = x^x$.
Then, let us define a function $p(x)$ such that:
$$p(x) = \frac {f(x+1)}{f(x)} - \frac {f(x)}{f(x-1)}$$
As the value of $x$ increases, $p(x)$ approaches $e$, or (I think),
$$\lim_{x\rightarrow \infty} p(x) = e$$
I have no idea why this occurs as I'm no advanced math student, but could someone explain the reason to me? I just found this out playing with a calculator.

REPLY [8 votes]: $\frac {f(x+1)}{f(x)} = (x+1)\left(\frac {x+1}x \right)^x$. Define $g(x) = (1+1/x)^x$.
Then $p(x) = (x+1)g(x+1) - xg(x)$.
It is well known that $\lim_{x \to \infty} g(x) = e$, but we need a more precise development of $g(x)$ as $x \to \infty$ :  
$g(x) = \exp(\log(g(x))) = \exp(x \log(1+x^{-1})) \\ = \exp(x (x^{-1} - x^{-2}/2 + O(x^{-3}))
\\ = \exp(1 - x^{-1}/2 + O(x^{-2})) = e - (e/2)x^{-1} + O(x^{-2})$.
Now we can get $xg(x) = xe - e/2 + O(x^{-1})$
And finally $p(x) = (x+1)e - e/2 - xe + e/2 + O(x^{-1}) = e + O(x^{-1})$

REPLY [2 votes]: You are right. Use the fact
$$\lim_{x\rightarrow\infty}\bigg(1+\frac{1}{x}\bigg)^x = e$$