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Unformatted text preview: lim ( x;y ) ! (0 ; 0) x p x 2 + y 2 does not exist. Dont just show calculations. Explain what you are doing. 1 Solution: Notice that lim ( x;y ) ! (0 ; 0) along the positive x axis x p x 2 + y 2 = lim x ! + x p x 2 + 0 2 = lim x ! + x x = lim x ! + 1 = 1 and also notice that lim ( x;y ) ! (0 ; 0) along the positive y axis x p x 2 + y 2 = lim y ! + p 2 + y 2 = lim y ! + 0 = 0 . Since we get di/erent values using di/erent paths of approach, we conclude that the indicated limit does not exist. 3) For the function f ( x; y; z ) = 1 p x 2 + y 2 , &nd f x and f y . Solution: First notice that we can write f as f ( x; y ) = & x 2 + y 2 & 1 = 2 . We thus have f x = & 1 2 & x 2 + y 2 & 3 = 2 (2 x ) = & x ( x 2 + y 2 ) 3 = 2 . Similarly, we obtain f y = & y ( x 2 + y 2 ) 3 = 2 . 2...
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 Fall '10
 Ellermeyer
 Math

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