#### Answer

$${\text{ The limit does not exist}}$$

#### Work Step by Step

$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{{e^x}}}{{8{x^5} - 3{x^4}}} \cr
& {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr
& {\text{evaluate the limit substituting }}0{\text{ for }}x{\text{ into the numerator and }} \cr
& {\text{denominator}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to 0} {e^x} = {e^x} = 1 \cr
& \mathop {\lim }\limits_{x \to 0} \left( {8{x^5} - 3{x^4}} \right) = 8{\left( 0 \right)^5} - 3{\left( 0 \right)^4} = 0 \cr
& {\text{Since the limits do not lead to the indeterminate form }}\frac{0}{0}{\text{ or }}\frac{{ \pm \infty }}{{ \pm \infty }} \cr
& {\text{we cannot apply the l'Hospital's rule}}. \cr
& {\text{By l'Hospital' rule}}{\text{, the limit does not exist}} \cr} $$