HW01-solutions - zakaria (mmz255) – HW01 – gilbert –...

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Unformatted text preview: zakaria (mmz255) – HW01 – gilbert – (55485) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Welcome to Quest. Print off this as- signment and bring it to lectures as well as discussion groups. 001 10.0 points Determine if lim x →− 1 parenleftBig 5 x 2 + 3 x 2 + 1 parenrightBig exists, and if it does, find its value. 1. limit = 8 2. limit does not exist 3. limit = 4 correct 4. limit = 5 5. limit = 3 Explanation: Set f ( x ) = 5 x 2 + 3 , g ( x ) = x 2 + 1 . Then lim x →− 1 f ( x ) = 8 , lim x →− 1 g ( x ) = 2 . Thus the limits for both the numerator and denominator exist and neither is zero; so L’Hospital’s rule does not apply. In fact, all we have to do is use properties of limits. For then we see that limit = 4 . 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. B and C only 2. C only 3. all of them 4. A and C only 5. A and B only 6. A only 7. B only correct 8. none of them Explanation: A. By properties of limits lim x → 1 f ( x ) g ( x ) = 0 · 0 = 0 , so this limit is not an indeterminate form. B. Since lim x → 1 = ∞· , this limit is an indeterminate form. C. By properties of limits lim x → 1 F ( x ) g ( x ) = 2 = 1 , so this limit is not an indeterminate form. 003 10.0 points zakaria (mmz255) – HW01 – gilbert – (55485) 2 Determine if lim x → sin − 1 (4 x ) tan − 1 (5 x ) exists, and if it does, find its value....
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This note was uploaded on 06/05/2011 for the course MATH 408 D taught by Professor Gilbert during the Spring '11 term at University of Texas at Austin.

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HW01-solutions - zakaria (mmz255) – HW01 – gilbert –...

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