Suppose I have a map $T(x_{0},y_{0}) = (x_{1}, y_{1})$ on a plane where $a(y)$ is some analytic function of $y$, and the dot $\dot{}$ is differentiation w.r.t. $y$:
$$x_{1} = x_{0} - 2\dot{a}(y_{1})$$ $$y_{1} = y_{0} + \frac{a(y_{0}) + a(y_{1})}{x_{0}}$$
$T$ preserves the area form $(x + \dot{a}(y))dx \wedge dy$ (and thus is symplectic / exact symplectic after an appropriate change of variables). Let us rescale $x \mapsto \frac{x}{\epsilon}$ where $\epsilon$ is a small parameter. Then the map in rescaled variables reads
$$x_{1} = x_{0} - 2 \epsilon \dot{a}(y_{1})$$ $$y_{1} = y_{0} + \epsilon \left(\frac{a(y_{0}) + a(y_{1})}{x_{0}}\right)$$
And it preserves accordingly the symplectic form $(x + \epsilon \dot{a}(y)) dx \wedge dy$. Since $\epsilon$ is small, let us expand the map in series of $\epsilon$ and write the above as $T_{\epsilon}$:
$$x_{1} = x_{0} - 2\epsilon \dot{a}(y_{0}) + O(\epsilon^{2})$$ $$y_{1} = y_{0} + \frac{2\epsilon a(y_{0})}{x_{0}} + O(\epsilon^{2})$$
What symplectic form, if any, does the map $T_{\epsilon}$ preserve? I am assuming that $T_{\epsilon}$ when expanded to all orders in $\epsilon$ should again preserve $(x + \epsilon \dot{a}(y)) dx \wedge dy$ since $T_{\epsilon}$ converges to $T$, but when $T_{\epsilon}$ is cut off at some finite order (like above), it does not preserve that form...
Would be grateful if someone sheds light on this!