Answer
$${\text{The limit does not exist}}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to - 5} \frac{{{x^3} - 3{x^2} + 4x - 1}}{{{x^2} - 25}} \cr
& {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr
& {\text{evaluate the limit substituting }} - 5{\text{ for }}x{\text{ into the numerator and }} \cr
& {\text{denominator}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to - 5} \left( {{x^3} - 3{x^2} + 4x - 1} \right) = {\left( { - 5} \right)^3} - 3{\left( { - 5} \right)^2} + 4\left( { - 5} \right) - 1 = - 221 \cr
& \mathop {\lim }\limits_{x \to - 5} 3x = {x^2} - 25 = {5^2} - 25 = 0 \cr
& {\text{Since the limits do not lead to the indeterminate form }}\frac{0}{0}{\text{ or }}\frac{{ \pm \infty }}{{ \pm \infty }} \cr
& {\text{we cannot apply the l'Hospital's rule}}. \cr
& {\text{By l'Hospital' rule}}{\text{, the limit does not exist}} \cr} $$
