<s>[INST] How I can refine the proof of $\mathop {\lim }\limits_{(x,y)\, \to \,(3, - 1)} \left( {{x^2} + {y^2} - 4x + 2y} \right) = - \,4$ To prove that $$\mathop {\lim }\limits_{(x,y)\, \to \,(3, - 1)} \left( {{x^2} + {y^2} - 4x + 2y} \right) =  - \,4$$  I followed the following process:
Because the hypothesis and the definition of limit, we have:
$\sqrt {{{(x - 3)}^2} + {{(y + 1)}^2}}  < \delta \,\,\,;\,\,\,\,\left| {x - 3} \right| < \delta \,\,;\,\,\,\left| {y + 1} \right| < \delta $
Then I have to find an upper bound of:
$\left| {{x^2} + {y^2} - 4x + 2y + 4} \right|$, I complete square like this:
$\begin{array}{c}
\left| {{x^2} + {y^2} - 4x + 2y + 4} \right| = \left| {{x^2} - 6x + 9 + {y^2} + 2y + 1 + 2x - 6} \right|\\
 = \left| {{{(x - 3)}^2} + {{(y + 1)}^2} + 2(x - 3)} \right|
\end{array}$
then using the triangle inequality and relate delta epsilon:
$\begin{array}{c}
\left| {{{(x - 3)}^2} + {{(y + 1)}^2} + 2(x - 3)} \right| \le \left| {{{(x - 3)}^2} + {{(y + 1)}^2}} \right| + 2\left| {x - 3} \right|\\
 < {\delta ^2} + 2\delta \\
 = {(\delta  + 1)^2} - 1\\
 < {(\delta  + 1)^2} = \varepsilon 
\end{array}$
but the latter involves that: $\delta  = \sqrt \varepsilon-1$
and this is consistent for $\epsilon>1$, and not for $\epsilon$ as small as desired
So how I can refine this demonstration? because every time I find more disconcerting.
 [/INST] From your last step, you want that: $\delta^2 + 2\delta < \epsilon$, but this means $(\delta + 1)^2 < 1 + \epsilon$, and taking square root $\delta + 1 < \sqrt{1 + \epsilon}$, so $\delta < -1 + \sqrt{1 + \epsilon}$. You can take $\delta = \dfrac{-1 + \sqrt{1 + \epsilon}}{2}$, and it does the job.
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