Find limit of function of two variables (if exist) Find $$
\lim_{(x,0)\to (\pi,0)}\frac{3sin^3(x+y)}{y^2+(x-\pi)^2}.
$$
My attempt was to show that that limit does not exist. But for any pairs(checked by me) of $(x_n,y_n)\to(\pi,0)$ this function tends to $0$ (when $n\to \infty$) (for example I checked $(x_n,y_n)=(\frac{1}{n}+\pi,\frac{1}{n}).$ I would be grateful for your hints and ideas.

I would expect the limit to exist and be zero. 


*

*Note that $\sin (t\pm\pi)=\pm\sin t$. So $\sin (x+y)=-\sin (x-\pi+y)$. 

*We also have, for small $t$, $|\sin t|\leq t$. 

*And, $(a+b)^3\leq 4a^3+4b^3$. 
Thus, for $(x,y)$ close to $(\pi,0)$, 
\begin{align}
\left|\frac{3\sin^3(x+y)}{y^2+(x-\pi)^2}\right|
&=\left|\frac{3\sin^3(x-\pi+y)}{y^2+(x-\pi)^2}\right|
\leq\left|\frac{3(x-\pi+y)^3}{y^2+(x-\pi)^2}\right|
\leq\left|\frac{12(x-\pi)^3+12y^3}{y^2+(x-\pi)^2}\right|\\ \ \\
&\leq\left|\frac{12(x-\pi)^3 }{y^2+(x-\pi)^2}\right|+\left|\frac{ 12y^3}{y^2+(x-\pi)^2}\right|\\ \ \\
&\leq12|x-\pi|+12|y|,
\end{align}
using that 
$$
\left|\frac{ (x-\pi)^2 }{y^2+(x-\pi)^2}\right|\leq1,\ \ \ \left|\frac{ y^2 }{y^2+(x-\pi)^2}\right|\leq1. 
$$
