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Unformatted text preview: the ones above would also give you 2/2. 14.2 # 14 (6 points): Evaluate the limit or determine that it does not exist. lim ( x,y ) → (0 , 0) x 2 + y 2 1 + y 2 . Solution: The function x 2 + y 2 1 + y 2 is continuous everywhere in the xyplane because it is a rational function whose denominator 1 + y 2 is never zero. The limit can be evaluated by substitution lim ( x,y ) → (0 , 0) x 2 + y 2 1 + y 2 = 1 = 0 . 14.2 # 34 (6 points): Show that lim ( x,y ) → (0 , 0) y 2 x 2 + y 2 does not exist. Solution : Let f ( x,y ) = y 2 x 2 + y 2 . Approaching (0 , 0) along the line y = mx , lim ( x,mx ) → (0 , 0) f ( x,mx ) = lim x → m 2 x 2 x 2 + m 2 x 2 = m 2 1 + m 2 . Since the limit of f ( x,y ) depends on the direction from which ( x,y ) approaches (0 , 0), then the limit does not exist....
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 Winter '08
 Helton
 Math, lim, #, vertical trace, paraboloid opening downwards

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