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cal9 - choi(kc24547 HW09 BERG(56525 This print-out should...

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choi (kc24547) – HW09 – BERG – (56525) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine if lim x → − 1 parenleftBig x 3 + 7 x 2 + x + 1 x 2 + 3 parenrightBig exists, and if it does, find its value. 1. limit = 5 4 2. limit does not exist 3. limit = 3 2 correct 4. limit = 6 5. limit = 5 Explanation: Both the limits for the numerator and de- nominator exist and the limit of the denom- inator is not equal zero. Thus L’Hospital’s rule does not apply. Now lim x → − 1 ( x 3 + 7 x 2 + x + 1) = 6 , while lim x → − 1 ( x 2 + 3) = 4 . Consequently, by Properties of limits, limit = 3 2 . keywords: 002 10.0 points When f, g, F and G are functions such that lim x 1 f ( x ) = 0 , lim x 1 g ( x ) = 0 , lim x 1 F ( x ) = 2 , lim x 1 G ( x ) = , which, if any, of A. lim x 1 F ( x ) g ( x ) , B. lim x 1 g ( x ) G ( x ) , C. lim x 1 f ( x ) g ( x ) , are indeterminate forms? 1. A only 2. B only correct 3. none of them 4. A and B only 5. B and C only 6. C only 7. all of them 8. A and C only Explanation: A. By properties of limits lim x 1 F ( x ) g ( x ) = 2 0 = 1 , so this limit is not an indeterminate form. B. Since lim x 1 = ∞ · 0 , this limit is an indeterminate form. C. By properties of limits lim x 1 f ( x ) g ( x ) = 0 · 0 = 0 , so this limit is not an indeterminate form. 003 10.0 points
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choi (kc24547) – HW09 – BERG – (56525) 2 Determine the value of lim x 0 f ( x ) g ( x ) when f ( x ) = e 3 x - 1 , g ( x ) = x 2 + 5 x . 1. limit = 3 2 2. limit = 2 5 3. limit = 2 3 4. limit = 3 5 correct 5. limit = 5 2 6. limit does not exist Explanation: Since f, g are differentiable functions such that lim x 0 f ( x ) = lim x 0 g ( x ) = 0 , L’Hospital’s Rule can be applied: lim x 0 f ( x ) g ( x ) = lim x 0 f ( x ) g ( x ) = lim x 0 3 e 3 x 2 x + 5 . Consequently, lim x 0 f ( x ) g ( x ) = 3 5 . 004 10.0 points Find the value of lim x 0 + 6 x - ln x 2 x . 1. limit = 2 2. limit = 6 3. limit = 0 4. limit = 3 5. none of the other answers 6. limit = correct 7. limit = -∞ 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 = -∞ . On the other hand, 6 x - ln x 2 x = 3 - ln x 2 x . Thus lim x 0 + 6 x - ln x 2 x = 3 + 0 , which is not an indeterminate form. In fact, since the second term is positive, we see that lim x 0 + 6 x - ln x 2 x = . 005 10.0 points Determine if lim x 0 sin 1 (2 x ) tan 1 (3 x ) exists, and if it does, find its value. 1. limit = 2 2. limit = 3 2 3. limit does not exist 4. limit = 0 5. limit = 2 3 correct
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choi (kc24547) – HW09 – BERG – (56525) 3 6. limit = 3 Explanation: Since the limit has the form 0 / 0, it is in- determinate and L’Hospital’s Rule can be ap- plied: lim x 0 f ( x ) g ( x ) = lim x 0 f ( x ) g ( x ) with f ( x ) = sin 1 (2 x ) , g ( x ) = tan 1 (3 x ) , and f ( x ) = 2 1 - 4 x 2 , g ( x ) = 3 1 + 9 x 2 . In this case, f ( x ) g ( x ) = 2 3 parenleftBig 1 + 9 x 2 1 - 4 x 2 parenrightBig -→ 2 3 as x 0. Consequently, the limit exists and limit = 2 3 .
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