test1review-solutions - suleimenov (bs26835) test1review...

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Unformatted text preview: 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 weeks 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 dont have to do all fifty problems! Have fun :-) 001 0.0 points Use LHospitals 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 LHospitals 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 LHospitals Rule to deter- mine lim x f ( x ) g ( x ) . In this way, we can use LHospitals 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 . suleimenov (bs26835) test1review rusin (55565) 2 Then lim x f ( x ) = = lim x g ( x ) , and by applying LHospitals 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 LHospitals 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 LHospitals Rule twice we see that lim x f ( x ) g ( x ) = ....
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test1review-solutions - suleimenov (bs26835) test1review...

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