Question

Show that the function f is continuous but not differentiable at the origin

Answer 1

For continuity, a common trick is to express f(x,y)=g(x,y)h(x,y)f(x,y)=g(x,y)h(x,y) where gg has limit 00 at (0,0)(0,0) and hh is bounded in a punctured neighborhood of (0,0)(0,0). This is easy here:
g(x,y)=x,h(x,y)=y2x2+y2
g(x,y)=x,h(x,y)=y2x2+y2
because it's obvious that 0≤h(x,y)≤10≤h(x,y)≤1 for all (x,y)≠(0,0)(x,y)≠(0,0).

Differentiability doesn't imply continuity of the partial derivatives; in some sense it's the other way round.

This function has partial derivatives at (0,0)(0,0) and both are zero:
limh→0f(0+h,0)−f(0,0)h=limh→01hh⋅02h2+02=0
limh→0f(0+h,0)−f(0,0)h=limh→01hh⋅02h2+02=0
and
limh→0f(0,0+h)−f(0,0)h=limh→01h0⋅h2h2+02=0
limh→0f(0,0+h)−f(0,0)h=limh→01h0⋅h2h2+02=0
So, if the function is differentiable at (0,0)(0,0), its differential must be zero. Can you go on?

Answer 2

Answer:

ixts8tydpudpcjpoxyt86e60f s