Evaluate $\lim\limits_{x \to 1}\dfrac{x-x^x}{1-x+\ln x}$. Problem
Evaluate $$\lim\limits_{x \to 1}\frac{x-x^x}{1-x+\ln x}$$.
Solution
Consider using Taylor's formula. Expand $x^x $ and $\ln x$ at $x=1$. We obtain
$$x^x=1+(x-1)+(x-1)^2+o((x-1)^2);$$
$$\ln x=(x-1)-\frac{1}{2}(x-1)^2+o((x-1)^2).$$
Therefore
\begin{align*}
\lim_{x \to 1}
\frac{x-x^x}{1-x+\ln x}&=\lim_{x \to 1}\frac{x-[1+(x-1)+(x-1)^2+o((x-1)^2)]}{1-x+[(x-1)-\frac{1}{2}(x-1)^2+o((x-1)^2)]}\\
&=\lim_{x \to 1}\frac{-(x-1)^2-o((x-1)^2)}{-\frac{1}{2}(x-1)^2+o((x-1)^2)}\\
&=2.
\end{align*}
Please correct me if I'm wrong! Can we solve it by L'Hospital's rule?

Why no?
$$\lim_{x\rightarrow1}\frac{x-x^x}{1-x+\ln{x}}=\lim_{x\rightarrow1}\frac{1-x^x(1+\ln{x})}{-1+\frac{1}{x}}=\lim_{x\rightarrow1}\frac{-x^x(1+\ln{x})^2-x^x\cdot\frac{1}{x}}{-\frac{1}{x^2}}=2$$
