Euler Limit of $\frac{x^x-x}{1-x+\ln(x)}$ Without L'Hopital

You mention allowing series expansions. This problem is trivial with them. $$\lim_{x\to 1}\dfrac{x^x-x}{1-x+\ln(x)}$$ So that I can use McLaurin series I already know, I am going to enforce the substitution $u = x-1$ $$=\lim_{u\to 0}\dfrac{(u+1)^{u+1}-u-1}{\ln(u+1)-u}$$ $$=\lim_{u\to 0}\dfrac{[1+u+u^2+O(u^3)]-u-1}{[u-\frac{u^2}{2}+O(u^3)]-u}$$ $$=\lim_{u\to 0}\dfrac{u^2+O(u^3)}{-\frac{u^2}{2}+O(u^3)}$$ $$=\lim_{u\to 0}\dfrac{u^2}{-\frac{u^2}{2}}$$ $$=\color{red}{-2}$$

$y=\ln(x)$ actually works

$$\lim_{x\to 1}\dfrac{x^x-x}{1-x+\ln(x)}=\lim_{y \to 0}\dfrac{e^{ye^y}-e^y}{1-e^{y}+y}=\lim_{y \to 0}e^{y}\dfrac{e^{y(e^y-1)}-1}{1-e^{y}+y}=\lim_{y \to 0}\dfrac{e^{y(e^y-1)}-1}{1-e^{y}+y}$$

Now, by the definition of the derivative $$\lim_{z \to 0}\frac{e^z-1}{z}=1$$

Replacing $z=y(e^y-1)$ you get $$\lim_{y \to 0}\frac{e^{y(e^y-1)}-1}{y(e^y-1)}=1$$

Therefore $$\lim_{y \to 0}\frac{e^{y(e^y-1)}-1}{1+y-e^y}=\lim_{y \to 0}\frac{e^{y(e^y-1)}-1}{y(e^y-1)}\frac{y(e^y-1)}{1+y-e^y}=\lim_{y \to 0}\frac{y(e^y-1)}{1+y-e^y}$$

This last limit is much easier and can be calculated easily, either with power series, or by calculating $$ \lim_{y \to 0}\frac{y(e^y-1)}{1+y-e^y}=\lim_{y \to 0}\frac{y(e^y-1)}{y^2}\frac{y^2}{1+y-e^y}=\lim_{y \to 0}\frac{e^y-1}{y}\frac{y^2}{1+y-e^y}=\lim_{y \to 0}\frac{y^2}{1+y-e^y} $$ which is pretty standard.

You can collect a factor $x$ from the numerator and then do $x=t+1$, so the limit becomes $$ \lim_{t\to0}(1+t)\frac{(t+1)^t-1}{\log(1+t)-t} $$ (surely Euler didn't use “ln”).

The factor $1+t$ contributes $1$, so we can concentrate on the fraction. The numerator is $e^{t\log(t+1)}-1$ and can be expanded like $$ 1+t\log(1+t)+o(t\log(1+t))-1=t^2+o(t^2) $$ whereas the denominator is $$ \log(1+t)-t=t-\frac{t^2}{2}+o(t^2)-t=-\frac{t^2}{2}+o(t^2) $$ so the limit is $-2$.