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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

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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:

$$ \lim_{(x, y) \to (0, \pi)} \frac{x^2 y \sin y}{\sin^2 x + (\pi - y)^2} = 0 $$

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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 $\left|\frac{x^2 (\pi-v) \sin v } {\sin^2x + v^2 }\right| \leqslant \frac{x^2 (\pi-v) |\sin v |} {\sin^2{x} } \underset{{(x,v) \to (0, 0 ) } }\longrightarrow 0.$

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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 $$\begin{align} \lim_{r \to 0 } \frac{r^2\cos^2\theta (\pi-r\sin\theta) \sin (\pi-r\sin\theta) } {\sin^2(r\cos\theta) + r^2\sin^2\theta } &=\lim_{r \to 0 } \frac{r^2\cos^2\theta (\pi-r\sin\theta) \sin (\pi-r\sin\theta) } {(r\cos\theta+O(r^3\cos^3\theta))^2 + r^2\sin^2\theta }\\ &=\lim_{r \to 0 } \frac{r^2\cos^2\theta (\pi-r\sin\theta) \sin (\pi-r\sin\theta) } {r^2\cos^2\theta+O(r^4\cos^4\theta) + r^2\sin^2\theta }\\ &=\lim_{r \to 0 } \frac{\cos^2\theta (\pi-r\sin\theta) \sin (\pi-r\sin\theta) } {1+O(r^2\cos^4\theta)}\\ &=\frac{\cos^2\theta \cdot \pi\cdot \sin \pi } {1}=0\\ \end{align}$$

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