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I'm unable to prove that

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

I tried with polar coordinates but I'm unable to reach a function that depends only on $\rho$

$$0\le\frac{\rho^3\cos^3(\theta)\sin(\theta)}{\rho^2\cos^2(\theta)+|\rho\sin(\theta)|}=\frac{\rho^2\cos^3(\theta)\sin(\theta)}{\rho\cos^2(\theta)+|\sin(\theta)|}\leq \dots ?$$

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Use the fact that $x^2 \leq x^2 + |y|$ and then apply the squeeze theorem.

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Hint. After reading your work, you may show that $$\frac{x^2y}{x^2+|y|}=\frac{\rho^2\cos^2(\theta)|\sin(\theta)|}{\rho\cos^2(\theta)+|\sin(\theta)|} \leq \rho^2$$ that is $$\rho^2\cos^2(\theta)|\sin(\theta)|\leq \rho^3\cos^2(\theta)+\rho^2|\sin(\theta)|.$$

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Hint: since $0\le|\sin(\theta)|,|\cos(\theta)|\le 1$ you have that $$0\le\left|\frac{\rho^2\cos^3(\theta)\sin(\theta)}{\rho\cos^2(\theta)+|\sin(\theta)|}\right| \le \frac{\rho^2}{\rho[1-\sin^2(\theta)]+|\sin(\theta)|}\le \begin{cases} \rho^2 & 1<\rho\\ \\ \rho & 0< \rho\le 1 \end{cases}$$ since $0\le 1-\sin^2(\theta)\le 1$ and $|\sin(\theta)|\le 1$.

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$|\frac {x^{2}y} {x^{2}+|y|}| \leq x^{2}$ because $x^{2}+|y| \geq |y|$.

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Hint: Use that $$x^2+|y|\geq 2|x||y|^{1/2}$$ so $$\frac{x^2|y|}{x^2+|y|}\le \frac{|x|^2|y|}{2|x||y|^{1/2}}=\frac{1}{2}|x||y|^{1/2}$$ and this tends to zero for $x,y$ tends to zero.

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