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March 24, 2017

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lim (x^3-1)/(x^3+2x^2y+xy^2-x^2-2xy-y^2)
(x,y)->(1,0)

my question is can you approach (1,0) with y=x and does that change the the limit to

lim f(x,x)
(x,x)->(1,1)

in which case i get 3/4 as the limit. and if this is not how i go about doing it can you point me in the right direction?

  • calculus - ,

    You cannot approach along y = x because that relationship does not apply at (1,0).

    Try approaching along y = 0

    Lim (x^3-1)/(x^3+2x^2y+xy^2-x^2-2xy-y^2)
    = Lim (x^3-1)/(x^3 -x^2)
    x->1
    = Lim 3x^2/(3x^2 -2x)
    x->1
    = 3

    I used L'Hopital's rule.

  • calculus - ,

    alright thanks, my teacher never showed us that if you set one of the variables equal to zero you can make it a one variable function. so say i approach x=0 and find the limit of that to be infinity. does the original function f(x,y) not exist?

  • calculus - ,

    or does x=0 not work because x does not equal zero on the point (1,0). cause my teacher was fairly lazy in class and used the point (0,0) for every example.

  • calculus - ,

    <<or does x=0 not work because x does not equal zero on the point (1,0)? >>
    It does not work because x is not 1 at y = 0, where they want the limit evaluated.

  • calculus - ,

    then what can i approach by as well cause i also tried y=x-1 and that was just a mess.

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