Arely Briggs

2022-01-25

I have the following multivariable function: $f\left(x,y\right)=x{y}^{2}$, and I must prove it is continuous at $\left(1,1\right)$. I have come to: $|{\left({\left(x-1\right)}^{2}+{\left(y-1\right)}^{2}\right)}^{\frac{3}{2}}-1|\le {\left({\left(x-1\right)}^{2}+{\left(y-1\right)}^{2}\right)}^{\frac{3}{2}}+1$ (Triangle inequality) $<{\delta }^{3}+1=ϵ$.
What can I do with this 1? Any tips are welcome and thank you in advance!

One can pose this problem by considering the graph of the function as follow.
From the definition of continuity the $\epsilon$-neighbourhood of the image $f\left(1,1\right)=1$ determines a couple of planes $z=1+\epsilon$ and $z=1-\epsilon$
The intersections of the planes $1±\epsilon$ with the graph of $f$ determine two hyperbolas
$\left(\begin{array}{c}x\\ y\\ 1±\epsilon \end{array}\right)$
with $x{y}^{2}=1±\epsilon$.
Also a little circle of radius $\delta$, which is the border of a $\delta$-neighbourhood centered at the preimage ${\left(1,1\right)}^{\mathrm{\top }}$, 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 $p={\left(1,1,1\right)}^{\mathrm{\top }}$ , which will be determined by considering that for the curve $C$ can be parametrized as $C\left(y\right)={\left(\frac{1+\epsilon }{{y}^{2}},y,1+\epsilon \right)}^{\mathrm{\top }}$ and the distance function
$d\left(p,C\left(y\right)\right)=\sqrt{{\left(\frac{1+\epsilon }{{y}^{2}}-1\right)}^{2}+{\left(y-1\right)}^{2}+{\epsilon }^{2}}.$

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