Consider the 1-forms given by:
$w=\dfrac{e^x}{x^2+y^2}\Big((x\cos y+y\sin y)dy+(x\sin y-y\cos y)dx\Big);$
$w_0=\dfrac{-ydx}{x^2+y^2}+\dfrac{xdy}{x^2+y^2}$.
I am trying to show that $\lim_{x^2+y^2\to 0}\sqrt{x^2+y^2}(w-w_0)=0$(the 1-form that vanishes everywhere). For this I decided to split the limit in two, one with respect to $dx$ and the other with respect to $dy$. They are as follow:
$\lim_{x^2+y^2\to 0}\sqrt{x^2+y^2}\dfrac{e^x}{x^2+y^2}\Big(x\sin y-y\cos y-\dfrac{x}{e^x}\Big)$;
$\lim_{x^2+y^2\to 0}\sqrt{x^2+y^2}\dfrac{e^x}{x^2+y^2}\Big(x\cos y+y\sin y-\dfrac{y}{e^x}\Big)$.
Both are very similar and my approach was to show that the values of $\dfrac{e^x}{x^2+y^2}\Big(x\sin y-y\cos y-\dfrac{x}{e^x}\Big)$ are bounded near the origin but I had no success. Once this is proved I can apply the squeeze principle to compute the limit. Any hints are appreciated.
Ted Shifrin's hint solves the question. For closure's sake I will write down his hint as it already solves the problem.
Consider the Taylor polynomial of degree 2 of the function $f(x,y)=e^x(xcosy+ysiny)-x$.