Answer
$$\frac{5}{7}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{5{e^x} - 5}}{{{x^3} - 8{x^2} + 7x}} \cr
& {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr
& {\text{evaluate the limit substituting }}0{\text{ for }}x{\text{ into the numerator and }} \cr
& {\text{denominator}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to 0} \ln \left( {5{e^x} - 5} \right) = 5{e^{\left( 0 \right)}} - 5 = 5 - 5 = 0 \cr
& \mathop {\lim }\limits_{x \to 0} \left( {{x^3} - 8{x^2} + 7x} \right) = {\left( 0 \right)^3} - 8{\left( 0 \right)^2} + 7\left( 0 \right) = 0 \cr
& \cr
& {\text{Since the limits of both numerator and denominator are 0}}{\text{, }} \cr
& {\text{l'Hospital's rule applies}}{\text{. Differentiating the numerator and}} \cr
& {\text{denominator}}{\text{}} \cr
& {\text{for }}5{e^x} - 5 \to {D_x}\left( {5{e^x} - 5} \right) = 5{e^x} \cr
& {\text{for }}x \to {D_x}\left( {{x^3} - 8{x^2} + 7x} \right) = 3{x^2} - 16x + 7 \cr
& {\text{then}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{5{e^x} - 5}}{{{x^3} - 8{x^2} + 7x}} = \mathop {\lim }\limits_{x \to 0} \frac{{5{e^x}}}{{3{x^2} - 16x + 7}} \cr
& {\text{Find the limit of the quotient of the derivatives}} \cr
& = \frac{{5{e^0}}}{{3{{\left( 0 \right)}^2} - 16\left( 0 \right) + 7}} \cr
& = \frac{5}{7} \cr
& {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{5{e^x} - 5}}{{{x^3} - 8{x^2} + 7x}} = \frac{5}{7} \cr} $$
