Explanation Procedure used: L’ Hospital’s rule: 1: Check whether the lim x → a f ( x ) g ( x ) is an indeterminate form 0 0 or ± ∞ ± ∞ . 2: If so obtain the derivative of f ( x ) and g ( x ) , separately. 3: Compute the lim x → a f ′ ( x ) g ′ ( x ) , if this limit exist then lim x → a f ( x ) g ( x ) = lim x → a f ′ ( x ) g ′ ( x ) . 4 If it does not exist repeat the step 2 and step 3. Calculation: Step 1: Obtain the value of lim x → 0 + x ln ( e x − 1 ) . lim x → 0 + x ln ( e x − 1 ) = lim x → 0 + ln ( e x − 1 ) 1 x = ln ( e 0 − 1 ) 1 0 = ln ( 0 ) ∞ = ∞ ∞ It is observed to be an indeterminate form. Thus, apply L’Hospital’s rule. Step 2: It is noticed that f ( x ) = ln ( e x − 1 ) and g ( x ) = 1 x Obtain the derivative of f ( x ) . f ′ ( x ) = d d x ( ln ( e x − 1 ) ) = 1 ( e x − 1 ) ( e x ) Obtain the derivative of g ( x ) . g ′ ( x ) = d d x ( 1 x ) = − 1 x 2 Step 3: Compute the value of lim x → 0 + f ′ ( x ) g ′ ( x )

Calculus With Applications, Books a la Carte Plus MyLab Math Package (11th Edition)

11th Edition

ISBN: 9780133886849

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Chapter 12.7, Problem 35E

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The value of limx→0+xln(ex−1) by applying the L’Hospital’s rule.

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Chapter 12.7, Problem 34E

Chapter 12.7, Problem 36E

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Calculus With Applications, Books a la Carte Plus MyLab Math Package (11th Edition)

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The value of lim x → 0 + x ln ( e x − 1 ) by applying the L’Hospital’s rule.

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The value of limx→0+xln(ex−1) by applying the L’Hospital’s rule.

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