Edward C. answered 06/23/15
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Caltech Grad for math tutoring: Algebra through Calculus
This limit does not exist.
If you hold z constant at 0 and approach the origin along the line y = 2x, then
f(x,y,z) = f(x,2x,0) = (x2 + 4x2 + 0) / (x2 - 4x2 - 0) = (5x2) / (-3x2) = -5/3
so the limit as (x,y,z)--->(0,0,0) along this line is -5/3
But if you approach the origin along the line y = 3x with (z = 0) then
f(x,y,z) = f(x,3x,0) = (x2 + 9x2 + 0) / (x2 - 9x2 - 0) = (10x2) / (-8x2) = -5/4
so the limit as (x,y,z)--->(0,0,0) along this line is -5/4
In order for the limit to exist it must have the same value regardless of which path you use to approach the limit point. Since these 2 paths approaching the origin have different limiting values the limit does not exist.