test1review-solutions

# test1review-solutions - suleimenov(bs26835 test1review...

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suleimenov (bs26835) – test1review – rusin – (55565) 1 This print-out should have 50 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. These questions are for you to practice with. You also need to work on this week’s homework assignment. Together, they cover all the topics I consider to be fair game for the exam on Tuesday. The test questions will be similar (but fewer, and NOT multiple-choice). Note that this homework is NOT required and NOT graded – it is for your practice only. Ob- viously you don’t have to do all fifty problems! Have fun :-) 001 0.0points Use L’Hospital’s Rule to determine which of the inequalitites A. e x < x 2 + 100, B. 100 x < e x , C. e 2 x > xe x + 100, holds for all large x . 1. none of them 2. all of them 3. C only correct 4. A only 5. A and B only 6. A and C only 7. B only 8. B and C only Explanation: The notion of limit at infinity tells us that if lim x → ∞ f ( x ) g ( x ) = , then f ( x ) g ( x ) > 1 for all large x . But then f ( x ) > g ( x ) holds for all large x so long as g ( x ) > 0 for large x , allowing us to multiply through the inequality by g ( x ). Similarly, if lim x → ∞ f ( x ) g ( x ) = 0 , then f ( x ) g ( x ) < 1 for all large x , so if g ( x ) > 0 for all large x , then f ( x ) < g ( x ) for all large x . On the other hand, if lim x → ∞ f ( x ) = = lim x → ∞ g ( x ) , then we can use L’Hospital’s Rule to deter- mine lim x → ∞ f ( x ) g ( x ) . Similarly, if lim x → ∞ f ( x ) = 0 = lim x → ∞ g ( x ) , then we can use L’Hospital’s Rule to deter- mine lim x → ∞ f ( x ) g ( x ) . In this way, we can use L’Hospital’s Rule to compare the rates of growth or decay of f ( x ) and g ( x ) when x → ∞ . For the three given inequalities, therefore, we have to choose appropriate f and g and make sure that g ( x ) > 0 for all large x . A. FALSE: set f ( x ) = e x , g ( x ) = x 2 + 100 .

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suleimenov (bs26835) – test1review – rusin – (55565) 2 Then lim x → ∞ f ( x ) = = lim x → ∞ g ( x ) , and by applying L’Hospital’s Rule twice we see that lim x → ∞ f ( x ) g ( x ) = . Thus the inequality e x > x 2 + 100 , not e x < x 2 + 100 , holds for all large x . B. FALSE: set f ( x ) = e x , g ( x ) = 100 x . Then lim x → ∞ f ( x ) = 0 = lim x → ∞ g ( x ) , and by applying L’Hospital’s Rule we see that lim x → ∞ f ( x ) g ( x ) = 0 . Thus the inequality 100 x > e x , not 100 x < e x , holds for all large x . C. TRUE: set f ( x ) = e 2 x , g ( x ) = xe x + 100 . Then lim x → ∞ f ( x ) = = lim x → ∞ g ( x ) , and by applying L’Hospital’s Rule twice we see that lim x → ∞ f ( x ) g ( x ) = . Thus the inequality e 2 x > xe x + 100 holds for all large x . keywords: 002 0.0points Find the value of lim x 0 + 4 x ln x 3 x . 1. limit = 3 2. limit = 0 3. limit = 4 4. limit = −∞ 5. limit = 4 3 6. limit = correct 7. none of the other answers Explanation: Let’s first check if the given limit is an indeterminate form. Now ln x is defined for x > 0 and the graph of ln x has a vertical asymptote at x = 0; in addition, lim x 0 + ln x = −∞ .
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