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