a. Explanation Given: The function f ( a ) = g ( a ) = 0 and g ′ ( a ) ≠ 0 . Calculation: Obtain the value of lim x → a f ( x ) g ( x ) . Consider the Taylor series for f ( x ) and g ( x ) as follows, f ( x ) = f ( a ) + f ′ ( a ) ( x − a ) + 1 2 f ″ ( a ) ( x − a ) 2 + ⋯ and g ( x ) = g ( a ) + g ′ ( a ) ( x − a ) + 1 2 g ″ ( a ) ( x − a ) 2 + ⋯ Substitute f ( x ) and g ( x ) in lim x → a f ( x ) g ( x ) . That is, lim x → a f ( x ) g ( x ) = f ( a ) + f ′ ( a ) ( x − a ) + 1 2 f ″ ( a ) ( x − a ) 2 + ⋯ g ( a ) + g ′ ( a ) ( x − a ) + 1 2 g ″ ( a ) ( x − a ) 2 + ⋯ = ( 0 ) + f ′ ( a ) ( x − a ) + 1 2 f ″ ( a ) ( x − a ) 2 + ⋯ ( 0 ) + g ′ ( a ) (</ b. To determine To evaluate: The value of lim x → a f ( x ) g ( x ) and show it consist two applications of L’Hopital’s Rule.

Calculus: Early Transcendentals (3rd Edition)

3rd Edition

ISBN: 9780134763644

Author: William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz

Publisher: PEARSON

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Chapter 11.4, Problem 83E

To determine

To evaluate: The value of limx→af(x)g(x).

To determine

To evaluate: The value of limx→af(x)g(x) and show it consist two applications of L’Hopital’s Rule.

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Chapter 11.4, Problem 81E

Chapter 11.4, Problem 84E

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Chapter 11 Solutions

Calculus: Early Transcendentals (3rd Edition)

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