Решите придел(lim), plz.

Решите задачу:

$\displaystyle \lim_{x \to \infty} (\frac{x-2}{x+1})^{2x-3}=(\frac{\infty}{\infty})^\infty=\lim_{x \to \infty}(\frac{x+1-3}{x+1})^{2x-3}=\\=[\lim_{x \to \infty}(1+\frac{1}{-\frac{x+1}{3}})^{-\frac{x+1}{3}}}]^{-\frac{3}{x+1}*(2x-3)}=e^{\displaystyle lim_{x \to \infty}\frac{9-6x}{x+1}}=\\=e^{\displaystyle lim_{x \to \infty}\frac{x(\frac{9}{x}-6)}{x(1+\frac{1}{x})}}=e^{-6}=\frac{1}{e^6}$