## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 12 - Sequences and Series - 12.7 L'Hospital's Rule - 12.7 Exercises - Page 659: 10

#### Answer

$$\frac{1}{4}$$

#### Work Step by Step

\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{x{e^{ - x}}}}{{2{e^{2x}} - 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} x{e^{ - x}} = \left( 0 \right){e^0} = 0 \cr & \mathop {\lim }\limits_{x \to 0} \left( {2{e^{2x}} - 2} \right) = 2{e^0} - 2 = 2 - 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 }}x{e^{ - x}} \to {D_x}\left( {x{e^{ - x}}} \right) = x{D_x}\left( {{e^{ - x}}} \right) + {e^{ - x}}{D_x}\left( x \right) = - x{e^{ - x}} + {e^{ - x}} \cr & {\text{for }}x \to {D_x}\left( {2{e^{2x}} - 2} \right) = 4{e^{2x}} \cr & {\text{then}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{x{e^{ - x}}}}{{2{e^{2x}} - 2}} = \mathop {\lim }\limits_{x \to 0} \frac{{ - x{e^{ - x}} + {e^{ - x}}}}{{4{e^{2x}}}} \cr & {\text{Find the limit of the quotient of the derivatives}} \cr & = \frac{{ - \left( 0 \right){e^{ - 0}} + {e^{ - 0}}}}{{4{e^{2\left( 0 \right)}}}} \cr & = \frac{1}{4} \cr & {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{x{e^{ - x}}}}{{2{e^{2x}} - 2}} = \frac{1}{4} \cr}

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