<s>[INST] 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
 [/INST] 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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