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