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