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# exam1solns - Math 221 Summer 2008 C Gmez o D Milovich Exam...

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Math 221 C. G´ omez Summer 2008 D. Milovich Exam 1 June 30, 2008 Name: Solutions Discussion Section Instructor (circle one): omez Milovich Problem Score 1 2 3 4 5 6 Total 28 10 20 10 20 12 100 This exam contains 9 pages, and 6 (multi- part) problems. Before you begin, please make sure all the pages are here. The last page of the exam contains formulas you may not need. If you decide to tear off that page, please do so carefully. No calculators, notes, or books are allowed. You must show all your work, and ex- plain your reasoning to receive credit for your answers. Be sure to check your answers whenever pos- sible. Good luck!

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Math 221 Exam 1 2 1. Evaluate the limit, if it exists. If it does not exist, explain why not. (a) [7 points] lim x →- 2 - 2 x - 4 x 2 + 2 x = lim x →- 2 - 2( x + 2) x ( x + 2) = lim x →- 2 - 2 x = - 2 - 2 = 1 (b) [7 points] lim x →- 2 ( x + 3) | x + 2 | x + 2 ( x + 3) | x + 2 | x + 2 = ( ( x + 3) - ( x +2) x +2 , if x + 2 < 0 ( x + 3) ( x +2 x +2 ) , if x + 2 > 0 = ( - ( x + 3) , if x < - 2 x + 3 , if x > - 2 Then we have to take the one-sided limits: lim x →- 2 - ( x + 3) | x + 2 | x + 2 = lim x →- 2 - - ( x + 3) = - ( - 2 + 3) = - 1 lim x →- 2 + ( x + 3) | x + 2 | x + 2 = lim x →- 2 + ( x + 3) = ( - 2 + 3) = 1 Then the limit does not exist because the one-sided limits don’t agree.
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exam1solns - Math 221 Summer 2008 C Gmez o D Milovich Exam...

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