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I'm pretty sure that

\begin{equation} \lim_{(x,y) \rightarrow (0,0)} \frac{x^4y}{x^2 + y^2} = 0, \end{equation}

but I'm having some trouble proving it.

The only technique I'm aware of that can be used to show indeterminate limits of $\geq 2$ variables exist is the Squeeze Theorem. I've tried applying it here (by assuming $|y| < 1$ and bounding the quantity of interest by $\pm\frac{x^4y}{x^2 + y^2}$), but I didn't get anywhere.

Any help is appreciated.

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Since $|xy|\leq \frac{x^2+y^2}{2}$ by the GM-QM inequality, you simply have: $$\left|\frac{x^4 y}{x^2+y^2}\right|\leq \frac{1}{2}|x|^3.$$

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$$ 0\le\left|\frac{x^4y}{x^2 + y^2}\right|= \left|\frac{x^4y}{\|(x,y)\|^2}\right|\le \frac{\|(x,y)\|^5}{\|(x,y)\|^2}= \|(x,y)\|^3. $$

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Use polar coordinates: $x = r\cos t$, $y = r\sin t\Rightarrow x^2 + y^2 = r^2$, and $yx^4 = r^5\sin t(\cos t)^4$. So: $yx^4/(x^2 + y^2) = r^3\sin t(\cos t)^4 \to 0$ as $r\to 0$.

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$$ \left|\frac{x^4y}{x^2+y^2}\right| \le \left|\frac{x^4y}{x^2}\right| = \left|x^2y\right| \to 0. $$

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The basic inequality here is $x^2\le x^2+y^2$, from which you get $$ \frac{|x|}{\sqrt{x^2+y^2}}\le 1 $$ with a similar one for $y$. So $$ \left|\frac{x^4y}{x^2+y^2}\right|\le|x^3|. $$

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