## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 12 - Sequences and Series - 12.7 L'Hospital's Rule - 12.7 Exercises - Page 659: 3

#### Answer

$$0$$

#### Work Step by Step

\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{{x^5} - 2{x^3} + 4{x^2}}}{{8{x^5} - 2{x^2} + 5x}} \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( {{x^5} - 2{x^3} + 4{x^2}} \right) = {\left( 0 \right)^5} - 2{\left( 0 \right)^3} + 4{\left( 0 \right)^2} = 0 \cr & \mathop {\lim }\limits_{x \to 0} \left( {8{x^5} - 2{x^2} + 5x} \right) = 8{\left( 0 \right)^5} - 2{\left( 0 \right)^2} + 5\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}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{{D_x}\left( {{x^5} - 2{x^3} + 4{x^2}} \right)}}{{{D_x}\left( {8{x^5} - 2{x^2} + 5x} \right)}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{5{x^4} - 6{x^2} + 8x}}{{40{x^4} - 4x + 5}} \cr & {\text{Find the limit of the quotient of the derivatives}} \cr & = \frac{{5{{\left( 0 \right)}^4} - 6{{\left( 0 \right)}^2} + 8\left( 0 \right)}}{{40{{\left( 0 \right)}^4} - 4\left( 0 \right) + 5}} \cr & = \frac{0}{5} \cr & = 0 \cr & {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{{x^5} - 2{x^3} + 4{x^2}}}{{8{x^5} - 2{x^2} + 5x}} = 0 \cr}

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