Lim стремится к 1 x^3-x^2-x+1/x^3-3x+2=

Решите задачу:

$\lim_{x \to 1} \dfrac{x^3 - x^2 - x + 1}{x^3 - 3x + 2 } = \lim_{x \to 1} \dfrac{x^2(x - 1) - (x - 1)}{x^3 - x^2 + x^2 - x - 2x + 2 } = \\ \\ \lim_{x \to 1} \dfrac{(x - 1)(x^2 - 1)}{x^2(x - 1) + x(x - 1) - 2(x - 1) } = \\ \\ \lim_{x \to 1} \dfrac{(x - 1)(x - 1)(x + 1)}{(x - 1)(x^2 + x - 2) } = \lim_{x \to 1} \dfrac{(x - 1)(x - 1)(x + 1)}{(x - 1)(x^2 -x + 2x - 2) } = \\ \\ \lim_{x \to 1} \dfrac{(x - 1)(x - 1)(x + 1)}{(x - 1)[x(x - 1) + 2(x - 1)] } =$

$\lim_{x \to 1} \dfrac{(x - 1)(x - 1)(x + 1)}{(x - 1)(x - 1)(x + 2) } = \lim_{x \to 1} \dfrac{x + 1}{x + 2} = \dfrac{1 + 1}{1 + 2} = \dfrac{2}{3}$