Check the continuity of the function
f: R^2to R at(0,0) , where f is defined by
f(x,y) = { 3x^2y/( x^2+y^2) , if (x,y)≠(0,0)
{ 3 , if (x,y)= (0,0)
If lim(x,y)→(0,0)f(x,y)=f(0,0)from any path then f is continuous.lim(x,y)→(0,0)f(x,y)=lim(x,y)→(0,0)3x2yx2+y2 (put y=mx)=limx→03mx3x2(1+m2)=limx→03mx1+m2=0≠f(0,0)=3Therefore, the given function is not continuous at (0,0).If \space lim_{(x,y)\to(0,0)}f(x,y)=f(0,0)\\ \text{from any path then f is continuous.}\\ lim_{(x,y)\to(0,0)}f(x,y)=lim_{(x,y)\to(0,0)}\frac{3x^2y}{x^2+y^2}\space ( put \space y=mx)\\ =lim_{x\to0}\frac{3mx^3}{x^2(1+m^2)}\\ =lim_{x\to0}\frac{3mx}{1+m^2}\\ =0\neq f(0,0)=3\\ \text{Therefore, the given function is not continuous at (0,0).}If lim(x,y)→(0,0)f(x,y)=f(0,0)from any path then f is continuous.lim(x,y)→(0,0)f(x,y)=lim(x,y)→(0,0)x2+y23x2y (put y=mx)=limx→0x2(1+m2)3mx3=limx→01+m23mx=0=f(0,0)=3Therefore, the given function is not continuous at (0,0).
