Given
"f(x,y) = \\frac{4x^2y }{4x^4+y^2}"
To evaluate
"\\lim_{(x,y)\\to(0,0)}f(x,y) = \\lim_{(x,y)\\to(0,0)}\\frac{4x^2y }{4x^4+y^2}"
Choose path passes through (0,0), let
"y=mx"
then the limit along y=mx given by
"\\begin{aligned}\n \\lim_{(x,y)\\to(0,0)}\\frac{4x^2y }{4x^4+y^2}&= \\lim_{x\\to0}\\frac{4mx^3 }{4x^4+m^2x^2}\\\\\n&= \\lim_{x\\to0}\\frac{4mx }{4x^2+m^2 }\\\\\n&= \\lim_{x\\to0}\\frac{0 }{0+m^2 }=0\n\\end{aligned}"
Since the limit does not depend on m , then limit exist and equal to 0
"\\lim_{(x,y)\\to(0,0)}f(x,y) =f(0,0)=0"
Then f(x,y ) is continuous at (0,0)
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