Question #217932
Find limit of
f(x,y)=(x^3y^3)/(x^3+y^3) at (x,y)=(0,0)
1
Expert's answer
2021-08-22T17:48:10-0400

lim(x,y)(0,0)x3y3x3+y3puty=mxlimx0x31+m3=0Thus, limit exists and is equal to 0.lim_{(x,y)\to (0,0)} \frac{x^3y^3}{x^3+y^3}\\ \text{put}y=mx\\ lim_{x\to 0} \frac{x^3}{1+m^3}\\ =0\\ \text{Thus, limit exists and is equal to 0.}


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