\(\displaystyle\lim_{x \to +\infty}\frac{x^x}{(x-1)^x}= \displaystyle\lim_{x \to +\infty}\left ( \frac{x}{x-1} \right )^x= \displaystyle\lim_{x \to +\infty}\left ( \frac{x-1}{x} \right )^{-x} \\=\displaystyle\lim_{x \to +\infty}\left ( 1+\frac{1}{-x} \right )^{-x}=e \)