Prove limit with $\varepsilon$-$\delta$ of a multivariable function

Question

I have to use the $\varepsilon$-$\delta$ definition to prove that $$\lim_{(x,y)\to(1,2)} x^2+y=3$$

So using the definition we have $\forall \varepsilon >0 \ \exists \delta>0$ such that $0<\sqrt{(x-1)^2+(y-2)^2}< \delta$ then $\mid x^2+y-3\mid< \varepsilon$ but how should I manipulate with the last expression?

Answer 1

You should use the fact that$$|x^2+y-3|=|x^2-1+y-2|\leqslant|x^2-1|+|y-2|.$$

Answer 2

Let $\epsilon > 0$. Then there exists a $\delta$ such that $3\delta + \delta^2 = \epsilon$, by surjectivity.

If $\sqrt{(x-1)^2 + (y-2)^2} < \delta$, then $$|x-1| = \sqrt{(x-1)^2} \leq \sqrt{(x-1)^2 + (y-2)^2} < \delta$$ similarly, $|y-2| < \delta$.

As $|x-1| < \delta$, we have $$|x^2-1^2| < |(1 \pm \delta)^2 - 1| < 2\delta + \delta^2$$

Thus, $|x^2+y-3| \leq |x^2-1^2|+|y-2| \leq 3\delta + \delta^2 = \epsilon$.

Answer 3

It is always easier to manipulate quantities which have $0$ as a limit.

I suggest using $\begin{cases}x=1+u\\y=2+v\end{cases}\quad$ with $(u,v)\to (0,0)$

The expression becomes

$|x^2+y-3|=|u^2+2u+1+2+v-3|=|u^2+2u+v|\le|u(u+2)|+|v|$

Since $u$ is small you can consider $|u|<1\implies |2+u|<3$

Thus selecting $\delta=\min(1,\varepsilon)$ you get $|u|<\delta,|v|<\delta\implies |x^2+y-3|<3\varepsilon+\varepsilon=4\varepsilon$

Note: do not complicate your life using euclidian norm, since $\sqrt{u^2+v^2}\le |u|+|v|<2\delta$, this is equivalent to taking $\delta'=\frac 12\delta$, but much easier to conclude.