Answer
$$\frac{{25}}{3}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{2{e^{5x}} - 25{x^2} - 10x - 2}}{{5{x^3}}} \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( {2{e^{5x}} - 25{x^2} - 10x - 2} \right) = 2{e^{5\left( 0 \right)}} - 25{\left( 0 \right)^2} - 10\left( 0 \right) - 2 = 2 - 2 = 0 \cr
& \mathop {\lim }\limits_{x \to 0} 5{x^3} = 5{\left( 0 \right)^3} = 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 the numerator}} \to {D_x}\left( {2{e^{5x}} - 25{x^2} - 10x - 2} \right) = 10{e^{5x}} - 50x - 10 \cr
& {\text{for }}5{x^3} \to {D_x}\left( {5{x^3}} \right) = 15{x^2} \cr
& {\text{then}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{2{e^{5x}} - 25{x^2} - 10x - 2}}{{5{x^3}}} = \mathop {\lim }\limits_{x \to 0} \frac{{10{e^{5x}} - 50x - 10}}{{15{x^2}}} \cr
& {\text{Find the limit of the quotient of the derivatives}} \cr
& = \frac{{10{e^{5\left( 0 \right)}} - 50\left( 0 \right) - 10}}{{15{{\left( 0 \right)}^2}}} = \frac{0}{0} \cr
& {\text{Since the limits of both numerator and denominator are 0}}{\text{, }} \cr
& {\text{l'Hospital's rule applies}}{\text{.}} \cr
& {\text{for }}10{e^{5x}} - 50x - 10 \to {D_x}\left( {10{e^{5x}} - 50x - 10} \right) = 50{e^{5x}} - 50 \cr
& {\text{for }}15{x^2} \to {D_x}\left( {15{x^2}} \right) = 30x \cr
& {\text{then}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{10{e^{5x}} - 50x - 10}}{{15{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{50{e^{5x}} - 50}}{{30x}} = \frac{{50{e^0} - 50}}{{30\left( 0 \right)}} = \frac{0}{0} \cr
& {\text{Since the limits of both numerator and denominator are 0}}{\text{, }} \cr
& {\text{l'Hospital's rule applies}}{\text{.}} \cr
& {\text{for }}50{e^{5x}} - 50 \to {D_x}\left( {50{e^{5x}} - 50} \right) = 250{e^x} \cr
& {\text{for }}30x \to {D_x}\left( {30x} \right) = 30 \cr
& {\text{then}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{50{e^{5x}} - 50}}{{30x}} = \mathop {\lim }\limits_{x \to 0} \frac{{250{e^x}}}{{30}} = \mathop {\lim }\limits_{x \to 0} \frac{{25{e^x}}}{3} \cr
& {\text{Find the limit of the quotient of the derivatives}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{25{e^x}}}{3} = \frac{{25{e^0}}}{3} = \frac{{25}}{3} \cr
& {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{2{e^{5x}} - 25{x^2} - 10x - 2}}{{5{x^3}}} = \frac{{25}}{3} \cr} $$
