Posted by Bryant on Tuesday, February 28, 2012 at 12:51pm.
lim (x^31)/(x^3+2x^2y+xy^2x^22xyy^2)
(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=x1 and that was a mess and i also tried y=mx which did nothing.

Calculus  Steve, Tuesday, February 28, 2012 at 1:22pm
Hmmm. . . Looks like a removable singularity to me.
x^3  1 = (x1)(x^2 + xy + y^2)
x^3+2x^2y+xy^2x^22xyy^2
= x(x+y)^2  (x+y)^2
= (x1)(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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