i.
We consider (a,b)=(0,0)
Now the four different paths through (0,0) are
y=xy=x2y=x3y=x4
ii
f(x,y)={(xβa)4(yβ)2(xβa)2(yβb)β0βif (x,y)ξ =(a,b)if (x,y)=(a,b)βBut(a,b)=(0,0)f(x,y)={x4y2x2yβ0βif (x,y)ξ =(0,0)if (x,y)=(0,0)βNow,lim(x,y)β(0,0)βf(x,y)=lim(x,y)β(0,0)βx4y2x2yβy=xlim(x)β(0)βx4βx2x2βxβ=0y=x2lim(x)β(0)βx4βx4x2βx2β=0.5y=x3lim(x)β(0)βx4βx6x2βx3β=0y=x4lim(x)β(0)βx4βx8x2βx4β=0
iii.
We will conclude that along every path chosen the limit of the function exist.
iv.
The function is continuous at the origin along every path chosen except along y=x2 7
v.
f(x,y)=x4+y2x2yβfnβ=βxβfβ=(x4+y2)2(x2+y2)β2xyβx2y(4x3)βfnβ(0,0)=Nonexistanefnβ=βyβfβ=(x4+y2)2(x4+y2)βx2βx2y(2y)βfnβ(0,0)=Nonexistane
vi.
The function is differentiable at every point except at (0,0)
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