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Unformatted text preview: MATH 137 Assignment 3  Solutions 1. (a) lim x 1 x 3 1  x 1  By the definition of the absolute value function,  x 1  = x 1 if x 1 ( x 1) if x < 1 Since we are taking the limit as x 1 , the lower case is applicable. Also, we can factor ( x 3 1) = ( x 1)( x 2 + x + 1), to get lim x 1 x 3 1  x 1  = lim x 1 ( x 1)( x 2 + x + 1) ( x 1) = lim x 1 ( x 2 + x + 1) = 3 (b) lim x 2 6 x 2 3 x 1 Plugging into the limit gives a form. So we must multiply the numerator and denominator by both of the conjugate expressions: lim x 2 6 x 2 3 x 1 = lim x 2 ( 6 x 2)( 6 x + 2)( 3 x + 1) ( 3 x 1)( 6 x + 2)( 3 x + 1) = lim x 2 (6 x 4)( 3 x + 1) (3 x 1)( 6 x + 2) = lim x 2 (2 x )( 3 x + 1) (2 x )( 6 x + 2) = lim x 2 3 x + 1 6 x + 2 = 1 2 (c) lim x  x + 1  x 1  x Using the definition of the absolute value function, near the point...
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This note was uploaded on 12/18/2010 for the course ECONOMICS 120 taught by Professor Mesta during the Spring '10 term at Wilfred Laurier University .
 Spring '10
 Mesta

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