Chapter 12 - Sequences and Series - 12.7 L'Hospital's Rule - 12.7 Exercises - Page 659: 25

$$\frac{1}{9}$$

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{1 + \frac{1}{3}x - {{\left( {1 + x} \right)}^{1/3}}}}{{{x^2}}} \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( {1 + \frac{1}{3}x - {{\left( {1 + x} \right)}^{1/3}}} \right) = 1 + \frac{1}{3}\left( 0 \right) - {\left( {1 + 0} \right)^{1/3}} = 0 \cr & \mathop {\lim }\limits_{x \to 0} {x^2} = {0^2} = 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 }}1 + \frac{1}{3}x - {\left( {1 + x} \right)^{1/3}} \to {D_x}\left( {1 + \frac{1}{3}x - {{\left( {1 + x} \right)}^{1/3}}} \right) = 0 + \frac{1}{3} - \frac{1}{3}{\left( {1 + x} \right)^{ - 2/3}} \cr & {\text{for }}{x^2} \to {D_x}\left( {{x^2}} \right) = 2x \cr & {\text{then}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{1 + \frac{1}{3}x - {{\left( {1 + x} \right)}^{1/3}}}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{3} - \frac{1}{3}{{\left( {1 + x} \right)}^{ - 2/3}}}}{{2x}} \cr & {\text{Find the limit of the quotient of the derivatives}} \cr & = \frac{{\frac{1}{3} - \frac{1}{3}{{\left( {1 + 0} \right)}^{ - 2/3}}}}{{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 }}\frac{1}{3} - \frac{1}{3}{\left( {1 + x} \right)^{ - 2/3}} \to {D_x}\left( {\frac{1}{3} - \frac{1}{3}{{\left( {1 + x} \right)}^{ - 2/3}}} \right) = \frac{2}{9}{\left( {x + 1} \right)^{ - 5/3}} \cr & {\text{for }}2x \to {D_x}\left( {2x} \right) = 2 \cr & {\text{then}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{3} - \frac{1}{3}{{\left( {1 + x} \right)}^{ - 2/3}}}}{{2x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{2}{9}{{\left( {x + 1} \right)}^{ - 5/3}}}}{2} \cr & {\text{Find the limit of the quotient of the derivatives}} \cr & = \frac{{\frac{2}{9}{{\left( {0 + 1} \right)}^{ - 5/3}}}}{2} = \frac{1}{9} \cr & {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{1 + \frac{1}{3}x - {{\left( {1 + x} \right)}^{1/3}}}}{{{x^2}}} = \frac{1}{9} \cr} $$