I have to formally prove $$\lim_{(x,y)\to (2,1)} x^2y=4$$ I am having a hard time getting from $|x^2y-4|$ to two evaluations $|x-2|< \delta_1$ and $|y-1|< \delta_2$. Does someone have some hints, because I am stuck, I have tried to add some values, but still not getting nowhere. I am trying to use the epsilon-delta definition for a proof: $$\forall \epsilon > 0 \exists \delta _1, \delta _2 > 0 : [(x,y) \neq (2,1), |x-2|< \delta _1, |y-1|< \delta _2] \Rightarrow |x^2y-4|<\epsilon$$
What could be next step from $|x^2y-4|=...$?
If $x,y\in\Bbb R$, then\begin{align}x^2y-4&=(x^2-4)(y-1)+x^2+4y-8\\&=(x^2-4)(y-1)+x^2-4+4y-4\\&=(x+2)(x-2)(y-1)+(x+2)(x-2)+4(y-1).\end{align}Now, suppose that $|x-2|<1$. Then $-1<x-2<1$ and therefore $1<x+2<3$; in particular, $|x+2|<3$, and therefore\begin{align}\bigl|x^2y-4\bigr|&=\bigl|(x+2)(x-2)(y-1)+(x+2)(x-2)+4(y-1)\bigr|\\&\leqslant3|x-2||y-1|+3|x-2|+4|y-1|.\end{align}Now, take $\varepsilon>0$. In order to have $\bigl|x^2y-4\bigr|<\varepsilon$, it is then enough that three things occur:
So, take $\delta>0$ such that $\delta\leqslant\frac{\sqrt\varepsilon}3$ and also that $\delta\leqslant\frac\varepsilon{12}$ (you can take $\delta=\min\left\{\frac{\sqrt\varepsilon}3,\frac\varepsilon{12}\right\}$). Then, if $\sqrt{(x-2)^2+(y-1)^2}<\delta$, you have that both numbers $|x-2|$ and $|y-1|$ are smaller than $\delta$ and it follows from the previous computations that then $\bigl|x^2y-4\bigr|<\varepsilon$.