Answer
$$4$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 1} \frac{{{x^3} + {x^2} - x - 1}}{{{x^2} - x}} \cr
& {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr
& {\text{evaluate the limit substituting }}1{\text{ for }}x{\text{ into the numerator and }} \cr
& {\text{denominator}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to 1} \left( {{x^3} + {x^2} - x - 1} \right) = {\left( 1 \right)^3} + {\left( 1 \right)^2} - \left( 1 \right) - 1 = 0 \cr
& \mathop {\lim }\limits_{x \to 1} \left( {{x^2} - x} \right) = {\left( 1 \right)^2} - \left( 1 \right) = 0 \cr
& \cr
& {\text{Since the limits of both numerator and denominator are 0}}{\text{, }} \cr
& {\text{l'Hospital's rule applies}}{\text{. differentiating the numerator and}} \cr
& {\text{denominator}} \cr
& = \mathop {\lim }\limits_{x \to 1} \frac{{{D_x}\left( {{x^3} + {x^2} - x - 1} \right)}}{{{D_x}\left( {{x^2} - x} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 1} \frac{{3{x^2} + 2x - 1}}{{2x - 1}} \cr
& {\text{Find the limit of the quotient of the derivatives}} \cr
& = \frac{{3{{\left( 1 \right)}^2} + 2\left( 1 \right) - 1}}{{2\left( 1 \right) - 1}} \cr
& = \frac{4}{1} = 4 \cr
& {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{{x^3} + {x^2} - x - 1}}{{{x^2} - x}} = 4 \cr} $$