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March 31, 2015

March 31, 2015

Posted by **Bryant** on Tuesday, February 28, 2012 at 12:51pm.

(x,y)->(1,0)

alright i approached this at y=0. and got the limit as 3. where else can i approach this from to see if the limit really exist which im hoping it doesn't. i tried y=x-1 and that was a mess and i also tried y=mx which did nothing.

- Calculus -
**Steve**, Tuesday, February 28, 2012 at 1:22pmHmmm. . . Looks like a removable singularity to me.

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

x^3+2x^2y+xy^2-x^2-2xy-y^2

= x(x+y)^2 - (x+y)^2

= (x-1)(x+y)^2

so, f(x,y) = (x^2+xy+y^2)/(x+y)^2 for x≠1

limit as (x,y) -> (1,0) = (1)/(1)^2 = 1

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