Show That Lim (x,y) => (0,0) (xy^2)/(x^2 + Y^2) Do

Anonymous

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Show that lim (x,y) => (0,0) (xy^2)/(x^2 + y^2) does not exist. Please show work on paper**

 

Answer No.1

MathTeacher

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F is not continuous at (0,0). The limit of f as (x,y) ->(0,0) does not exist. Consider the pair of paths: take (x,y) -> (0,0) along y = x lim x^3/(x^2 + x^4) = 0. x->0 take (x,y) ->(0,0) along y = v|x| lim x|x|/(x + x) = 1 and x->0+ lim x|x|/(x + x) = -1. x->0- Feel free to contact me. If you have any doubts Thanks :)

Answer No.2

TinyPro6

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F is not continuous at (0,0). The limit of f as (x,y) ->(0,0) does not exist. Consider (x,y) -> (0,0) along y = x lim x3/(x2 + x4) = 0. x->0 take (x,y) ->(0,0) along y = v|x| lim x|x|/(x + x) = 1 and x->0+ lim x|x|/(x + x) = -1.

Answer No.3

Anonymous

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Watch http://www.youtube.com/watch?v=q9xIdF33ql8.....u will understand.........

Answer No.4

Anonymous

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Lim (x,y) => (0,0) (xy^2)/(x^2 + y^2) exists and =0 suppose y=sin x and as per L hospital rule, lim (x,y) => (0,0) (xy^2)/(x^2 + y^2) = lim (x,y) => (0,0), x sin2x /(x^2 + sin 2x), take derivative, then lim (x,y) => (0,0), sin2x + x sin 2x /(2x + sin 2x), take again derivative, then lim (x,y) => (0,0), 2sin2x + 2 cos 2x/(2 + 2cos2x), =lim (x,y) => (0,0), sin2x + cos 2x/(1 + cos2x), as x and y approaches to 0, numerartor approaches to 1 and denominator to 2 hence it will be (1/2). If question is lim (x,y) => (0,0) (xy^2)/(x^2 - y^2) , then it does not exist. please check.

Answer No.5

Anonymous

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If (x,y) -> (0,0) along y = x lim x3/(x2 + x4) = 0. x->0 take (x,y) ->(0,0) along y = v|x| lim x|x|/(x + x) = 1 and x->0+ lim x|x|/(x + x) = -1.

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