Exam3_2 10

# Exam3_2 10 - 2 ln x 3 x 6 = lim x →∞ 2 x 3 = 0...

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Taylor, Douglas – Exam 3 – Due: Dec 5 2007, 1:00 am – Inst: JEGilbert 10 3. limit = 9 4. none of the other answers 5. limit = 6. limit = -∞ Explanation: Use of L’Hospital’s Rule is suggested, Set f ( x ) = (ln x ) 2 , g ( x ) = 3 x + 6 ln x. Then f, g have derivatives of all orders and lim x →∞ f ( x ) = , lim x →∞ g ( x ) = . Thus L’Hospital’s Rule applies: lim x →∞ f ( x ) g ( x ) = lim x →∞ f 0 ( x ) g 0 ( x ) . But f 0 ( x ) = 2 ln x x , g 0 ( x ) = 3 + 6 x , so lim x →∞ f 0 ( x ) g 0 ( x ) = lim x →∞ 2 ln x 3 x + 6 . We need to apply L’Hospital once again, for then lim x →∞
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Unformatted text preview: 2 ln x 3 x + 6 = lim x →∞ 2 x 3 = 0 . Consequently, the limit exists and lim x →∞ (ln x ) 2 3 x + 6 ln x = 0 . keywords: L’Hospital’s Rule, log function, limit, indeterminate form, indeterminate quo-tient 018 (part 1 of 1) 10 points Determine which one of the following could be the graph of f ( x ) = x 2-2 4-x 2 (dashed lines indicate asymptotes). 1. 2. 3. 4....
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