Give a $\delta - \epsilon$ proof of $\displaystyle \lim_{(x,y)\rightarrow(2,3)}(2x^3+3y^2-2y)$.

Let $z=f(x,y)=(2x^3+3y^2-2y)$, then $f(2,3)=37$.

Therefore, if (1) $0 < \sqrt{(x-2)^2+(y-3)^2} < \delta \le 1$, then (2) $0 < |2x^3+3y^2-2y-37| < \epsilon$.

From (1), we have (3) $0 < |x-2| < \delta\le 1$ and $0 < |y-3| < \delta\le 1$.

Moving on to (2), $0 < |2x^3+3y^2-2y-37| = |2x^3+3y^2-18y+27+16y-64| = |2(x^3-8)+3(y-3)^2+16(y-3)|$

$=|2(x-2)(x^2+2x+4)+(y-3)(y+7)|\le 2|x-2||x^2+2x+4|+|y-3||3y+7| < \epsilon$

From (3), we have $0 < |x-2|< 1$, so $ 1 < x < 3 $, therefore $7 < |x^2+2x+4| < 19$.

Similarly from (3), $0 < |y-3| < 1$, so $ 2< y < 4$, therefore $13 < |3y+7| < 19$.

So, $0 < 2|x-2||x^2+2x+4|+|y-3||3y+7| \le 38|x-2|+19|y-3| < \epsilon$

Therefore, $\delta=\min\{1,\frac{\epsilon}{57}\}. $