(I am a newbie to this site so apologies if it's been asked before but I can't find it)
I was manually messing about and noticed that
$$f(x)=\frac{x^x}{{(x-1)}^{(x-1)}}$$
has a slope that converges to $e$. (I literally calculated, like, $5^5/4^4 = 2.73...$, etc.).
I want to show this in a formal proof, so I took the derivative which is
$$f'(x)=\frac{x^{x}}{{(x-1)}^{(x-1)}}(\ln(x) - \ln(x-1))$$
and tried to rearrange/simplify/distribute/write in exponential form to show that this must be equal to $e$, but am completely stuck. Is there a formal proof to show that the above is equal to $e$? I cannot find anything online. Thank you so much!
Note that $$ f'(x)=\frac{x^x}{(x-1)^{x-1}} (\ln x-\ln(x-1)) = \left(\frac{x}{x-1}\right)^{x-1} \cdot x(\ln x-\ln(x-1)). $$ We have that $$ \lim_{x \to \infty} \left(\frac{x}{x-1}\right)^{x-1} = \lim_{x \to \infty} \left( 1+\frac{1}{x-1}\right)^{x-1} = \lim_{y\to \infty} \left( 1+\frac{1}{y}\right)^y= e, $$ using the substitution $y=x-1$. To compute the limit of the other part, note that $\ln x-\ln(x-1)$ converges to $0$, while $x$ converges to $\infty$, and so we can apply the L'Hopital rule, as follows: $$\begin{align*} \lim_{x \to \infty} x(\ln x-\ln(x-1)) &= \lim_{x \to \infty} \frac{\ln x-\ln(x-1)}{1/x} \\ &= \lim_{x \to \infty} \frac{1/x-1/(x-1)}{-1/x^2} \\ &= \lim_{x \to \infty} \frac{x}{x-1} \\ &=\lim_{x \to \infty} \frac{1}{1-1/x} = 1. \end{align*}$$ This gives us that $\lim_{x \to \infty} f'(x)=e$.