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notes001 (83) - 3 y ) 3 , ±nd @[email protected] , @[email protected] , @ 2 [email protected]@y ,...

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October 4, 2010 NAME____________________________ 1) By considering di/erent paths of approach, show that lim ( x;y ) ! (0 ; 0) x 4 x 4 + y 2 does not exist. Solution: First we try approaching (0 ; 0) along the y axis (where x = 0 ). This gives us lim ( x;y ) ! (0 ; 0) w ith x =0 x 4 x 4 + y 2 = lim y ! 0 0 4 0 4 + y 2 = 0 . Now let us try approaching (0 ; 0) along the x axis (where y = 0 ). This gives us lim ( x;y ) ! (0 ; 0) w ith y =0 x 4 x 4 + y 2 = lim x ! 0 x 4 x 4 + 0 2 = 1 . Since two di/erent results were obtained for di/erent paths of approach, we conclude that lim ( x;y ) ! (0 ; 0) x 4 x 4 + y 2 does not exist. 2) For the function f ( x; y ) = (2 x
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Unformatted text preview: 3 y ) 3 , ±nd @[email protected] , @[email protected] , @ 2 [email protected]@y , and @ 2 [email protected]@x . Do not just write down answers. You must include intermediate steps in your calculations. Solution: @f @x = 3 (2 x & 3 y ) 2 (2) = 6 (2 x & 3 y ) 2 . @f @y = 3 (2 x & 3 y ) 2 ( & 3) = & 9 (2 x & 3 y ) 2 . @ 2 f @[email protected] = @ @x & & 9 (2 x & 3 y ) 2 ± = & 18 (2 x & 3 y ) (2) = & 36 (2 x & 3 y ) . @ 2 f @[email protected] = @ @y & 6 (2 x & 3 y ) 2 ± = 12 (2 x & 3 y ) ( & 3) = & 36 (2 x & 3 y ) . 1...
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This note was uploaded on 02/05/2012 for the course MATH 2203 taught by Professor Ellermeyer during the Fall '10 term at FIU.

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