Multivariable limit - Two variables $ \lim_{(x,y) \to (0, \pi ) } \frac{x^2 y \sin y } {\sin^2x + (\pi - y )^2 }$ How can I calculate the following limit and show that it equals $0$:
$$ \lim_{(x,y) \to (0, \pi ) } \frac{x^2 y \sin y } {\sin^2x + (\pi - y )^2 }$$
Thanks in advance
 A: Let's prove the limit using the definition. Fix $\varepsilon > 0$. We have:
$$
\left| \frac{x^2 y \sin y}{\sin^2 x + (\pi - y)^2} \right| \le \left| \frac{x^2 y \sin y}{\sin^2 x} \right| = \left| \frac{x}{\sin x} \right|^2 \cdot \left|y \sin y\right|
$$
We know that $\lim_{x \to 0}\frac{x}{\sin x} = 1$ and $\lim_{y \to \pi} y \sin y = 0$. Therefore, we can pick a neighborhood of $(0, \pi)$ so that:
$$
\left| \frac{x}{\sin x} \right|^2 < 1 + \varepsilon, \ \left|y \sin y\right| < \varepsilon
$$
Thus:
$$
\left| \frac{x^2 y \sin y}{\sin^2 x + (\pi - y)^2} \right| \le \varepsilon(1 + \varepsilon)
$$
Since our choice of $\varepsilon$ was arbitrary, we conclude:
$$
\lim_{(x, y) \to (0, \pi)} \frac{x^2 y \sin y}{\sin^2 x + (\pi - y)^2} = 0
$$
A: Introduce new variable $v= \pi-y$. Then $\lim\limits_{(x,y) \to (0, \pi ) } \frac{x^2 y \sin y } {\sin^2x + (\pi - y )^2 }=\vert v= \pi-y \vert=\lim\limits_{(x,v) \to (0, 0 ) } \frac{x^2 (\pi-v) \sin v } {\sin^2x + v^2 }.$ Desired result can be obtained from the estimate $\left|\frac{x^2 (\pi-v) \sin v } {\sin^2x + v^2 }\right| \leqslant \frac{x^2 (\pi-v) |\sin v |} {\sin^2{x} } \underset{{(x,v) \to (0, 0 ) } }\longrightarrow 0.$
A: First I would change coordinates to $(x,z)$ where $z=\pi-y$. The limit becomes
$$\lim_{(x,z) \to (0, 0 ) } \frac{x^2 (\pi-z) \sin (\pi-z) } {\sin^2x + z^2 }$$
which we can evaluate by changing to polar coordinates and using the fact that near $0$, $\sin x=x+O(x^3)$. If you aren't familiar with big O notation, you can read about it on Wikipedia. This gives us
$$\begin{align}
\lim_{r \to 0 } \frac{r^2\cos^2\theta (\pi-r\sin\theta) \sin (\pi-r\sin\theta) } {\sin^2(r\cos\theta) + r^2\sin^2\theta } &=\lim_{r \to 0 } \frac{r^2\cos^2\theta (\pi-r\sin\theta) \sin (\pi-r\sin\theta) } {(r\cos\theta+O(r^3\cos^3\theta))^2 + r^2\sin^2\theta }\\
&=\lim_{r \to 0 } \frac{r^2\cos^2\theta (\pi-r\sin\theta) \sin (\pi-r\sin\theta) } {r^2\cos^2\theta+O(r^4\cos^4\theta) + r^2\sin^2\theta }\\
&=\lim_{r \to 0 } \frac{\cos^2\theta (\pi-r\sin\theta) \sin (\pi-r\sin\theta) } {1+O(r^2\cos^4\theta)}\\
&=\frac{\cos^2\theta \cdot \pi\cdot \sin \pi } {1}=0\\
\end{align}$$
