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hw2s - -x 2(2-x x = lim x → 2(2-x(2 x(2-x x = lim x → 2...

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MAT 1330, Fall 2011 Assignment 2 Due WED October 5, at the beginning of class. Late assignments will not be accepted; nor will unstapled assignments. Instructor (circle one): Robert Smith? Jason Levy Olga Vassilieva Catalin Rada DGD (circle one): 1 2 3 4 Student Name Student Number By signing below, you declare that this work was your own and that you have not copied from any other individual or other source. Signature Question 1. Find lim x 0 x 4 + sin 3 x 2011 . Answer: 0 Give one sequence to support your claim. Four terms are enough. x n f ( x n ) 0.5 0.062996019 0.1 0.000246951 0.05 0.00008056 0.01 0.000014925 Question 2. Does the limit lim t 0 2011 t 3 - t + 9 exist? Answer: Yes Justify your answer in at most two lines without using sequences of numerical values for t . Answer: lim t 0 2011 t 3 - t + 9 = lim t 0 2011 t (3 + t + 9) (3 - t + 9)(3 + t + 9) = lim t 0 2011 t (3 + t + 9) 9 - t - 9 = lim t 0 2011(3 + t + 9) - 1 = - 12066 . 1
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Question 3. Let F ( x ) = 4 - x 2 | 2 - x | + | x | . a) Find lim x 2 + F ( x ). lim x 2 + 4 - x 2 | 2 - x | +
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Unformatted text preview: -x 2-(2-x ) + x = lim x → 2 + (2-x )(2 + x )-(2-x ) + x = lim x → 2 + 2 + x-1 + x =-2 b) Find lim x → 2-F ( x ). lim x → 2-4-x 2 | 2-x | + | x | = lim x → 2-4-x 2 2-x + x = lim x → 2-(2-x )(2 + x ) 2-x + x = lim x → 2-2 + x + x = 6 c) Does lim x → 2 F ( x ) exist? Answer: No Justify your answer: lim x → 2 + F ( x ) 6 = lim x → 2-F ( x ) = ⇒ The limit at 2 does not exist. Question 4. Consider the function f ( x ) = 2011 x 2 . Use the definition of the derivative to compute f (2). Answer: f (2) = lim h → f (2 + h )-f (2) h = lim h → 2011 (2+ h ) 2-2011 2 2 h = lim h → 2011 { 2 2-(2 + h ) 2 } h 2 2 (2 + h ) 2 = lim h → 2011(2-2-h )(2 + 2 + h ) h 2 2 (2 + h ) 2 = lim h → 2011(-1)(4 + h ) 2 2 (2 + h ) 2 =-2011 4 . 2...
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