Consider
(x,y)→(0,0)limx2+y36x2y3 Suppose ε>0
∣∣x2+y36x2y3∣∣=6∣∣x2+y3y3∣∣x2 Note that
∣∣x2+y3y3∣∣≤1,x2≤x2+y2=δ2 So
6∣∣x2+y3y3∣∣x2<6⋅1⋅δ2 If we choose δ=6ε then
6∣∣x2+y3y3∣∣x2<6⋅1⋅δ2=ε
∣f(x,y)−0∣<ε,if x2+y2<δ Hence
(x,y)→(0,0)limx2+y36x2y3=0
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