Limit of two variables function by defintion I have to prove by definition that
$$\lim_{(x,y)\to(3,\pi)}x^{2}\sin(y)=0$$
Since
$$|f(x,y)-0|=|x^{2}|\cdot|\sin(y)|$$
and
$$|\sin(y)|\leq1$$
then
$$|f(x,y)-0|\leq|x^{2}|$$
Can I say that since
$$|x|\leq\sqrt{x^{2}+y^{2}}$$
so
$$|x^{2}|=|x|^{2}\leq x^{2}+y^{2}$$
and therefore
$$|x|^{2}\leq(x-3)^{2}+(y-\pi)^{2}?$$
I'm not sure about the last inequality, but it's the only way I found to try to write $\delta$ using $\epsilon$.
EDIT
I tried to change the variables:
$$x=\bar{x}+3,\quad y=\bar{y}+\pi$$
so, I have the limit
$$\lim_{(\bar{x},\bar{y})\to(0,0)}(\bar{x}+3)^{2}\sin(\bar{y}+\pi)$$
and, still
$$|f(\bar{x},\bar{y})-0|\leq|(\bar{x}+3)^2|$$
but now I must to find $\delta$ such that
$$||(\bar{x},\bar{y})-(0,0)||<\delta$$
implies on
$$|f(\bar{x},\bar{y})-0|<\epsilon$$
Since
$$||(\bar{x},\bar{y})-(0,0)||=\sqrt{\bar{x}^{2}+\bar{y}^{2}}$$
then
$$\sqrt{\bar{x}^{2}+\bar{y}^{2}}<\delta\Rightarrow \bar{x}^{2}+\bar{y}^{2}<\delta^{2}$$
and, therefore
$$|\bar{x}|^{2}=\bar{x}^{2}\leq \bar{x}^{2}+\bar{y}^{2}<\delta^{2}\Rightarrow |\bar{x}|^{2}<\delta^{2}$$
and so
$$|\bar{x}|<\delta$$
Hence,
$$|(\bar{x}+3)^{2}|=|\bar{x}^{2}+6\bar{x}+9|\leq|\bar{x}^{2}|+6|\bar{x}|+9<\delta^{2}+6\delta+9=(\delta+3)^{2}$$
Then, if I take
$$\epsilon=(\delta+3)^{2}$$
the limit is proved.
I want to know if it's right.
 A: You are asked about the following limit: $\ \lim\limits_{(x,y)\to(3,\pi)}f(x,y)$

*

*we have $x\to 3$ and $y\to \pi$ so we want to compare $f(x,y)$ to $f(3,\pi)=9\sin(\pi)=0$
Our mathematical toolbox works better with limits in $0$ or $\infty$ so when we have a limit in "random" points, we often shift the function to study it around $0$.

*

*let set $\begin{cases}x=3+u &&u\to 0\\y=\pi+v&&v\to 0\end{cases}$
Notice $(u,v)\to(0,0)\implies (x,y)\to(3,\pi)$
You were on the right track but I feel that using $\bar x,\bar y$ just add some more confusion (also it is heavy to write), prefer the use of $(u,v)$ or $(h,k)$ or whatever seems familiar to you, using always the same letters will trigger math reflexes.

*

*we now have to study $\bigg|f(x,y)-f(3,\pi)\bigg|=\bigg|(3+u)^2\sin(\pi+v)-0\bigg|=\bigg|(3+u)^2\sin(v)\bigg|$
Since $\sin(\pi+v)=-\sin(v)$
Now you recognize $\sin(v)\to 0$ when $v\to 0$, for that we can use the inequality $|\sin(v)|\le |v|$
On the other hand $(3+u)^2$ does not really interests us, let just say $u\to 0$ so it is small, without loss of generality we can assume $|u|<1$ and then $(3+u)^2<16$
Note: look at the end the choice of delta, $|u|<1$ is just convenient and not a limitation.

*

*finally we arrived to $\bigg|f(x,y)-f(3,\pi)\bigg|\le 16|v|\to 0$
Now regarding the epsilon-delta definition of the limit, remember that norms in $\mathbb R^2$ are equivalent.
In particular using one of the $3$ standard ones $\begin{cases}\lVert(u,v)\lVert_1=|u|+|v|\\\\\lVert (u,v)\lVert_2=\sqrt{u^2+v^2}\\\\\lVert(u,v)\lVert_\infty=\max(|u|,|v|)\\\end{cases}$ is up to you.
Select the one which is more convenient for the problem at hand

*

*since $\begin{cases}|u|\le |u|+|v|&&|u|\le\sqrt{u^2+v^2}\\|v|\le|u|+|v|&&|v|\le\sqrt{u^2+v^2}\end{cases}\implies \lVert(u,v)\lVert_\infty\le\lVert(u,v)\lVert_1\le2\lVert(u,v)\lVert_2$


*

*also $\sqrt{u^2+v^2}\le\sqrt{u^2}+\sqrt{v^2}=|u|+|v|\implies \lVert(u,v)\lVert_2\le\lVert(u,v)\lVert_1\le2\lVert(u,v)\lVert_\infty$

In our case since we have to bound $|u|$ and $|v|$ then select $$\delta=\min(1,\frac{\epsilon}{16})\text{ and }\lVert(u,v)\lVert_\infty<\delta$$

We have our result since $\begin{cases}|u|\le\max(|u|,|v|)<\delta\le 1\\|v|\le\max(|u|,|v|)<\delta\le \frac{\epsilon}{16}\end{cases}$
Notice that we do not have to do particularly confusing operations with $x-3$ and $y-\pi$ and square roots. The fact that we used $u,v$ and norm equivalence leads to a natural solution for the problem.
A: I'd do it this way:
We want:
$|x^2\sin(y)-0|<\epsilon$
We know that
$|x^2\sin(y)-0|$
$=|x^2\sin(y)|$
$= |x^2||\sin(y)|$
$= |(x-3+3)^2||\sin(y-\pi+\pi)|$
$ = |(x-3)+3|^2|\sin(y-\pi+\pi)|$
$ \le (|x-3|+|3|)^2|\sin((y-\pi)+\pi)|$
$\sin(a+b) = \sin(a)\cos(b) + \cos(a)\sin(b)$
so
$ = (|x-3|+3)^2|\sin(y-\pi)|$
So here I use that $|sin(a)|\le|a|$
$ \le (|x-3|+3)^2|y-\pi|$
We force $|x-3|<1$.
So then
$ (|x-3|+3)^2|y-\pi|$
$ < (1+3)^2|y-\pi|$
$ = 16|y-\pi|$
We want $16|y-\pi|$ to be $< \epsilon$
ie: we want $|y-\pi| < \frac{\epsilon}{16}$
Now we can do our proof.
Give an $\epsilon>0$ we choose $\delta = \min\{1,\frac{\epsilon}{16}\}$
We've already shown that
$|x^2\sin(y)-0|$
$\le (|x-3|+3)^2|y-\pi|$
Now since $|x-3|<\delta$ and $|y-\pi|<\delta$
$(|x-3|+3)^2|y-\pi|$
$< (\delta+3)^2\delta$
$\le (1+3)^2\delta$
$\le 16\delta$
$\le 16(\frac{\epsilon}{16})$
$= \epsilon$
