Lecture-9 - x 2 f ( x ) exists. Answer: lim x 2 + f ( x ) =...

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Math 137, Lecture 9, Friday September 26, 2008 Topic: Limits Exercises for Student: 1. For f ( x ) = 3 x | x - 2 | x - 2 evaluate lim x 2 + f ( x ) and lim x 2 - f ( x ). Hence, deduce if lim
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Unformatted text preview: x 2 f ( x ) exists. Answer: lim x 2 + f ( x ) = 6, lim x 2-f ( x ) =-6 limit does not exist. 2. Evaluate lim x 1 x-x 2 1- x . Answer: 3 1...
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This note was uploaded on 02/14/2012 for the course MATH 137 taught by Professor Speziale during the Spring '08 term at Waterloo.

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