Answer
$$\eqalign{
& \left( {\text{a}} \right){\text{ The L'Hopital's rule does not apply twice}}{\text{.}} \cr
& \left( {\text{b}} \right){\text{ }}2 \cr} $$
Work Step by Step
$$\eqalign{
& \left( {\text{a}} \right)\mathop {\lim }\limits_{x \to 1} \frac{{{x^3} - {x^2} + x - 1}}{{{x^3} - {x^2}}} \cr
& {\text{Evaluating the limit}} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{{x^3} - {x^2} + x - 1}}{{{x^3} - {x^2}}} = \mathop {\lim }\limits_{x \to 1} \frac{{3{x^2} - 2x + 1}}{{3{x^2} - 2x}} \cr
& {\text{The limit }}\mathop {\lim }\limits_{x \to 1} \frac{{3{x^2} - 2x + 1}}{{3{x^2} - 2x}}{\text{ is:}} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{3{x^2} - 2x + 1}}{{3{x^2} - 2x}} = \frac{{3{{\left( 1 \right)}^2} - 2\left( 1 \right) + 1}}{{3{{\left( 1 \right)}^2} - 2\left( 1 \right)}} = \frac{2}{1} = 2 \cr
& {\text{Then, the L'Hopital's rule does not apply}}{\text{. }} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{3{x^2} - 2x + 1}}{{3{x^2} - 2x}} = \mathop {\lim }\limits_{x \to 1} \frac{{6x - 2}}{{6x - 2}}{\text{ should not be done}}{\text{.}} \cr
& \cr
& \left( {\text{b}} \right) \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{{x^3} - {x^2} + x - 1}}{{{x^3} - {x^2}}} = \mathop {\lim }\limits_{x \to 1} \frac{{3{x^2} - 2x + 1}}{{3{x^2} - 2x}} = 2 \cr
& {\text{The correct limit is 2}} \cr} $$
