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\begin{equation} f(x,y) = \left\{ \begin{array}{l l} \sqrt{x(y-1)^2}, \quad x(y-1)^2 \geq 0\\ 0, \quad x(y-1)^2 <0 \\ \end{array} \right\} \end{equation}

It is clear to me how to prove where $x(y-1)^2\neq0$, it is the border that is giving me some trouble.

I believe I was able to prove it is continuous when y=1, through the following:

$$ |\sqrt{x(y-1)^2}|=\sqrt{x(y-1)^2}=\sqrt{x}.|y-1| $$

if $ \varepsilon \le 1 \implies 0 \leq \sqrt{x} \le \sqrt{a+1} $

$$ \sqrt{x}.|y-1| \le \sqrt{a+1}.|y-1| \le \sqrt{a+1}.||(x-a),(y-1)||<\sqrt{a+1}.\varepsilon $$

$$ \forall\delta>0,0<\varepsilon\le\frac{\delta}{\sqrt{a+1}}:||(x,y)-(a,1)||<\varepsilon \implies |f(x,y)-0|<\delta, a\in R^+_{0} $$

I'm just not sure how to go about replicating this for when $(x,y) \rightarrow (0,a), a \in R$

Thank you in advance.

André Fradinho
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    You can write this function as the composition of $(x,y)\to x(y-1)^2$ and $r(t)=\sqrt t$ if $t \geq 0$, $r(t)=0$ if $t<0$ – Thomas Dec 31 '21 at 13:20
  • @Thomas Thank you for your answer. However, I am failling to understand how this helps me prove that the system is continuous in the entire set of Real numbers, could you please elaborate a little? – André Fradinho Dec 31 '21 at 16:51
  • If $f$ and $r$ are continuous, then so is $r\circ f$. – Thomas Jan 01 '22 at 12:09
  • @Thomas Oh, of course, that makes perfect sense, thank you. But I was looking for the classic definition of continuity, so I'll keep the thread open. – André Fradinho Jan 01 '22 at 12:56
  • @Thomas: That looks like an answer. – Lee Mosher Jan 01 '22 at 14:53

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