Answer
$$ - 2$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{{e^{2x}} - 1}}{{5{x^2} - x}} \cr
& {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr
& {\text{evaluate the limit substituting }}0{\text{ for }}x{\text{ into the numerator and }} \cr
& {\text{denominator}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to 0} \left( {{e^{2x}} - 1} \right) = {e^{2\left( 0 \right)}} - 1 = 0 \cr
& \mathop {\lim }\limits_{x \to 0} \left( {5{x^2} - x} \right) = 5{\left( 0 \right)^2} - \left( 0 \right) = 0 \cr
& \cr
& {\text{Since the limits of both numerator and denominator are 0}}{\text{, }} \cr
& {\text{l'Hospital's rule applies}}{\text{. Differentiating the numerator and}} \cr
& {\text{denominator}}{\text{.}} \cr
& {\text{for }}{e^{2x}} - 1 \to {D_x}\left( {{e^{2x}} - 1} \right) = 2{e^{2x}} \cr
& {\text{for }}x \to {D_x}\left( {5{x^2} - x} \right) = 10x - 1 \cr
& {\text{then}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{e^{2x}} - 1}}{{5{x^2} - x}} = \mathop {\lim }\limits_{x \to 0} \frac{{2{e^{2x}}}}{{10x - 1}} \cr
& {\text{Find the limit of the quotient of the derivatives}} \cr
& = \frac{{2{e^{2\left( 0 \right)}}}}{{10\left( 0 \right) - 1}} \cr
& = \frac{2}{{ - 1}} \cr
& = - 2 \cr
& {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{e^{2x}} - 1}}{{5{x^2} - x}} = - 2 \cr} $$
