<s>[INST] Show using the definition of limit that $\lim_{ (x,y)\to(0,0)}\frac{ (1-\cos(xy))\sin y}{(x^2+y^2) }= 0$ can you help me with this excercise.
Show using the definition of limit that
$$\lim_{ (x,y)\to(0,0)}\frac{ (1-\cos(xy))\sin y}{(x^2+y^2) }= 0$$

Definition of limit:
$\lim_{(x,y)\to(a,b)} f(x,y) =L$ if and only if for every $\epsilon >0$ exist $\delta>0$ such that if $\sqrt{(x-a)^2+(y-b)^2}<\delta$ then $|f(x,y)-L|<\epsilon$.

Hi 
I´ve tried this, 
For taylor series.
Given $\epsilon>0$, find $\delta>0$ such that if
$$\sqrt{x^2+y^2}<\delta$$ then $$\bigg|\frac{(1-\cos xy)\sin y}{x^2+y^2}\bigg|<\epsilon$$
\begin{align*}
\bigg|\frac{(1-\cos xy)\sin y}{x^2+y^2}\bigg| & = \bigg|\frac{(1-1-\frac{x^2y^2}{2}+\frac{x^4y^4}{24})(y-\frac{y^3}{6}+\frac{y^5}{120})}{x^2+y^2}\bigg|\\
& =\bigg|\frac{(-\frac{x^2y^2}{2}+\frac{x^4y^4}{24})(y-\frac{y^3}{6}+\frac{y^5}{120})}{x^2+y^2}\bigg|
\end{align*}
Then what do I do?
 [/INST] We need only use the inequality $\sin x<x$ for $x>0$ along with the trigonometric identity $\sin^2 x=\frac{1-\cos 2x}{2}$.  
Then, we can write 
$$\begin{align}
|1-\cos xy|&=|2\sin^2(xy/2)|\\\\
&\le\frac12(xy)^2\\\\
&\le \frac14(x^2+y^2)^2
\end{align}$$
along with 
$$\begin{align}
|\sin y| &\le |y|\\\\
&\le(x^2+y^2)^{1/2}
\end{align}$$
Therefore, we have
$$\begin{align}
\left|\frac{(1-\cos xy)\sin y}{x^2+y^2}-0\right|&\le\frac{\frac14(x^2+y^2)^{5/2}}{x^2+y^2} \\\\
&=\frac14(x^2+y^2)^{3/2}\\\\
&<\epsilon
\end{align}$$
whenever $(x^2+y^2)^{1/2}<\delta = (4\epsilon)^{1/3}$.
</s>