Answer
$$0$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 1} \frac{{{x^4} - {x^3} - 3{x^2} + 5x - 2}}{{{x^3} + {x^2} - 5x + 3}} \cr
& {\text{Evaluate the limit}} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{{x^4} - {x^3} - 3{x^2} + 5x - 2}}{{{x^3} + {x^2} - 5x + 3}} = \frac{{{1^4} - {1^3} - 3{{\left( 1 \right)}^2} + 5\left( 1 \right) - 2}}{{{1^3} + {1^2} - 5\left( 1 \right) + 3}} = \frac{0}{0} \cr
& {\text{Use L'hopitals Rule}} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{{x^4} - {x^3} - 3{x^2} + 5x - 2}}{{{x^3} + {x^2} - 5x + 3}} = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{d}{{dx}}\left[ {{x^4} - {x^3} - 3{x^2} + 5x - 2} \right]}}{{\frac{d}{{dx}}\left[ {{x^3} + {x^2} - 5x + 3} \right]}} \cr
& {\text{ }} = \mathop {\lim }\limits_{x \to 1} \frac{{4{x^3} - 3{x^2} - 6x + 5}}{{3{x^2} + 2x - 5}} \cr
& {\text{Evaluate the limit}} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{4{x^3} - 3{x^2} - 6x + 5}}{{3{x^2} + 2x - 5}} = \frac{0}{0} \cr
& {\text{Use L'hopitals Rule}} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{4{x^3} - 3{x^2} - 6x + 5}}{{3{x^2} + 2x - 5}} = \mathop {\lim }\limits_{x \to 1} \frac{{12{x^2} - 6x - 6}}{{6x + 2}} \cr
& {\text{Evaluate the limit}} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{12{x^2} - 6x - 6}}{{6x + 2}} = \frac{{12 - 6 - 6}}{{6\left( 1 \right) + 2}} = 0 \cr
& {\text{Therefore}} \cr
& \mathop {\lim }\limits_{x \to 1} \frac{{{x^4} - {x^3} - 3{x^2} + 5x - 2}}{{{x^3} + {x^2} - 5x + 3}} = 0 \cr} $$
