Step 3Similarly, approaching along the y-axis yields a limit equal to 0. Since these two limits are the same, we will examine another approach path. Approach (0,0) along the curve y = x2.When x is positive, we havelim(x, y)(0,0)xyx2+ y2=limx 0x33x2+ x4=limx 0x2$$1+x2.When x is negative, we havelim(x, y)(0,0)xyx2+ y2=limx 0x33x2+ x4=limx 0x2$$1+x2.Step 4This limit evaluates aslimx 0x21+ x2=1=.