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Unformatted text preview: lim x 16 1 x + 4 = 1 8 . Hence, it follows that lim x 16 x4 x16 = 1 8 . 3. Determine each of the following limits, or explain why they dont exist. (a) lim x 2 + 3( x + 3)  x + 2  x + 2 (b) lim x 23( x + 3)  x + 2  x + 2 1 (a) Answer: Since were interested in x >2, we know that x + 2 > 0, which means that  x + 2  = x + 2 . Hence, lim x 2 + 3( x + 3)  x + 2  x + 2 = lim x 2 + 3( x + 3) x + 2 x + 2 = 3 1 1 = 3 . (b) Answer: In this part, were interested in x < 2, which means that, for the values of x we care about, x + 2 < 0. Therefore,  x + 2  =( x + 2) , so we have lim x 23( x + 3)  x + 2  x + 2 = lim x 23( x + 3)( x + 2) x + 2 = 3 1 (1) =3 . 2...
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This note was uploaded on 01/18/2012 for the course MATH 2250 taught by Professor Chestkofsky during the Fall '08 term at University of Georgia Athens.
 Fall '08
 CHESTKOFSKY
 Math, Calculus

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