Chapter 12 - Sequences and Series - Chapter Review - Review Exercises - Page 662: 63

$$\frac{9}{2}$$

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \left( {\frac{{{e^{3x}}}}{{{x^2}}} - \frac{1}{{{x^2}}} - \frac{3}{x}} \right) \cr & {\text{The common denominator is }}{x^2},{\text{ then}} \cr & \frac{{{e^{3x}}}}{{{x^2}}} - \frac{1}{{{x^2}}} - \frac{3}{x} = \frac{{{e^{3x}} - 1 - 3x}}{{{x^2}}} \cr & \mathop {\lim }\limits_{x \to 0} \left( {\frac{{{e^{3x}}}}{{{x^2}}} - \frac{1}{{{x^2}}} - \frac{3}{x}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{{e^{3x}} - 1 - 3x}}{{{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^{3x}} - 1 - 3x}}{{{x^2}}}} \right) = \frac{{{e^{3\left( 0 \right)}} - 1 - 3\left( 0 \right)}}{{{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^{3x}} - 1 - 3x \to {D_x}\left( {{e^{3x}} - 1 - 3x} \right) = 3{e^{3x}} - 3 \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^{3x}} - 1 - 3x}}{{{x^2}}}} \right) = \mathop {\lim }\limits_{x \to 0} \frac{{3{e^{3x}} - 3}}{{2x}} \cr & {\text{Find the limit of the quotient of the derivatives}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{3{e^{3x}} - 3}}{{2x}} = \frac{{3{e^0} - 3}}{{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 }}3{e^{3x}} - 3 \to {D_x}\left( {3{e^{3x}} - 3} \right) = 9{e^{3x}} \cr & {\text{for }}2x \to {D_x}\left( {2x} \right) = 2 \cr & {\text{then}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{3{e^{3x}} - 3}}{{2x}} = \mathop {\lim }\limits_{x \to 0} \frac{{9{e^{3x}}}}{2} \cr & {\text{Find the limit of the quotient of the derivatives}} \cr & = \frac{{9{e^{3\left( 0 \right)}}}}{2} = \frac{9}{2} \cr & {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr & \mathop {\lim }\limits_{x \to 0} \left( {\frac{{{e^{3x}}}}{{{x^2}}} - \frac{1}{{{x^2}}} - \frac{3}{x}} \right) = \frac{9}{2} \cr} $$