Assignment 3 Solutions - MATH 137 Assignment 3 - Solutions...

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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 .

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Assignment 3 Solutions - MATH 137 Assignment 3 - Solutions...

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