I have the following multivariable function: f(x,y) = xy^2, and

Arely Briggs

Arely Briggs

Answered question

2022-01-25

I have the following multivariable function: f(x,y)=xy2, and I must prove it is continuous at (1,1). I have come to: |((x1)2+(y1)2)321|((x1)2+(y1)2)32+1 (Triangle inequality) <δ3+1=ϵ.
What can I do with this 1? Any tips are welcome and thank you in advance!

Answer & Explanation

Addisyn Thompson

Addisyn Thompson

Beginner2022-01-26Added 16 answers

One can pose this problem by considering the graph of the function as follow.
From the definition of continuity the ε-neighbourhood of the image f(1,1)=1 determines a couple of planes z=1+ε and z=1ε
The intersections of the planes 1±ε with the graph of f determine two hyperbolas
(xy1±ε)
with xy2=1±ε.
Also a little circle of radius δ, which is the border of a δ-neighbourhood centered at the preimage (1,1), when is transformed gives a closed curve C viewed in the graph of f.
Then there is a point in the curve C which is the farest away from p=(1,1,1) , which will be determined by considering that for the curve C can be parametrized as C(y)=(1+εy2,y,1+ε) and the distance function
d(p,C(y))=(1+εy21)2+(y1)2+ε2.

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