Using $\varepsilon$-$\delta$, prove that $\lim\limits_{(x,y) \rightarrow (3,2)} (xy^2+\frac{3}{x})=13$ I started working on factoring $|xy^2+\frac{3}{x}-13|$ and got to $|xy^2-12-\frac{x-3}{x}|$. I have to get to $(y-2)$ somehow, but am confused how to get that from $xy^2-12$. Can anyone give even a slight hint? :)

Let $\varepsilon>0$ be fixed.
We can write
\begin{align}
\big|xy^2+\frac 3x-13\big| &= \big|xy^2-3\cdot2^2+\frac3x-1\big| =\\[1.5ex]
&= \big|(x-3)\,4+x(y^2-2^2)+\frac {3-x}x\big|\leq\\[1.5ex]
&\leq \big|x-3\big|\,4+\big|x\big|\cdot\big|y-2\big|\cdot\big|y+2\big|+\frac{\big|3-x\big|}{\big|x\big|}\leq\ldots\\[1.5ex]
\end{align}
$\Big[$Because $(x,y)$ converges to $(3,2)$, we can suppose that $2\leq x\leq 4,\quad 1\leq y\leq 3$, i.e. $(x,y)$ belongs to the closed square $S\Big((3,2),1\Big)$ centered in$(3,2)$ with half-side $1\Big]$
\begin{align}
\phantom{ll}\ldots&\leq\big|x-3\big|\,4+4\,\big|y-2\big|\,5+\frac12\,\big|3-x\big|=\\[1.5ex]
&=\frac92\,\big|x-3\big|+20\,\big|y-2\big|\leq\\[1.5ex]
\end{align}
\begin{align}
&\leq 20\Big(\big|x-3\big|+\big|y-2\big|\Big)\leq\ldots\\[1.5ex]
\end{align}
$\bigg[$ By using that $\;a,b\in\mathbb R \implies a+b\leq\sqrt2\sqrt{a^2+b^2}\bigg]$
\begin{align}
\phantom{l}\ldots&\leq 20\,\sqrt2\,\sqrt{\big|x-3\big|^2+\big|y-2\big|^2}\leq\varepsilon
\end{align}
when
$$ (x,y)\in B\Big((3,2), \frac{\varepsilon}{20\sqrt2}\Big)\cap S\Big((3,2),1\Big), $$
where $B\Big((\xi,\eta),r\Big)$ denote the closed disk centered in $(\xi,\eta)$ and with radius $r$.
Hence a possible choice for $\delta$ is
$$\delta=\min\Big(\frac\varepsilon{20\sqrt2},1\Big). $$
