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I can't figure out how to calculate this limit (or prove it does not exist)

$$ \lim_{(x,y) \to(0,0)} \frac{x^2y}{x^2+y^4} $$

I've tried with restrictions on $y=mx$ and curves of the form $y=x^n$. The limit should not exist but even with polar coordinates I can't figure it out

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Using polar: $\lvert\dfrac{r^3\cos^2\theta\sin\theta}{r^2(\cos^2\theta+r^2\sin^2\theta)}\rvert=\lvert\dfrac{r\cos^2\theta \sin\theta}{\cos^2\theta +r^2\sin^2\theta}\rvert\le\lvert\dfrac {r\cos^2\theta \sin\theta}{\cos^2\theta}\rvert=\lvert r\sin\theta\rvert\to0$, if $\theta\neq\dfrac{k\pi}2$. But it's easy to see the limit is $0$ when $\theta =\dfrac {k\pi}2$.

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    $\begingroup$ This is what I got but what if $\theta = \pi / 2 $ does it create problems? $\endgroup$ – DanieleMS Jul 19 at 11:26
  • $\begingroup$ I don't think so. See my edit. $\endgroup$ – Chris Custer Jul 19 at 11:45
  • $\begingroup$ So cool thanks I had not thought of that! $\endgroup$ – DanieleMS Jul 19 at 11:53
  • $\begingroup$ If $\theta=\pi/2$ then I believe everything after the $\leq$ is undefined, but the left side of the $\leq$ is zero so you already have the desired result in that case. $\endgroup$ – David K Jul 19 at 17:10
  • $\begingroup$ @DavidK good catch. I overlooked that. Plus I now see that I didn't need polar. $\endgroup$ – Chris Custer Jul 19 at 19:57
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If $(x,y) \neq (0,0)$, then we have \begin{align} \left| \dfrac{x^2y}{x^2 + y^4} \right| &= \left| \dfrac{x^2}{x^2 + y^4} \right| \cdot |y| \\ &\leq 1 \cdot |y| \\ &= |y| \end{align} From here it's easy to give an $\varepsilon$-$\delta$ argument for why the limit is $0$.

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    $\begingroup$ Nice answer! So simple! $\endgroup$ – quasi Jul 20 at 1:29
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Let $f(x,y)={\large{\frac{x^2y}{x^2+y^4}}}$.

Let $x^2+y^2=r^2$, with $0 < r \le 1$.

If $x\ne 0$, then \begin{align*} |f(x,y)|&=\left|\frac{x^2y}{x^2+y^4}\right|\\[4pt] &\le\left|\frac{x^2y}{x^2+x^2y^4}\right|\;\;\;\;\;\text{[since $x^2\le r^2\le 1$]}\\[4pt] &=\left|\frac{y}{1+y^4}\right|\\[4pt] &\le |y|\\[4pt] &\le r\\[4pt] \end{align*} and if $x=0$, then $y\ne 0$, so $$ f(x,y)=\frac{0}{y^4}=0 \qquad\qquad\qquad\qquad\qquad\;\;\; $$

In either case, we have $|f(x,y)|\le r$.

Letting $r$ approach zero from above, it follows that $$ \lim_{(x,y)\to (0,0)}f(x,y)=0 \qquad\qquad\qquad\qquad\qquad\;\;\; $$

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Just take the limit along the curve $x^2y=x^2+y^4$, or, solving for $x$,

$$ x= \sqrt{\frac{y^4}{y-1}} $$

If $(x,y)$ is on that curve, then $f(x,y)=1$, so the limit does not exist (if it existed, it should be 0 for what you've already concluded)

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    $\begingroup$ Good try but it doesn't work. If $y\le 0$, the equation $x^2y=x^2+y^4$ is impossible unless $(x,y)=(0,0)$. And if $y > 0$, then as $y$ approaches zero from the right, $y-1$ becomes negative. $\endgroup$ – quasi Jul 19 at 13:20
  • $\begingroup$ In any case, as has been shown in other answers, the limit is, in fact, equal to $0$. $\endgroup$ – quasi Jul 20 at 1:32

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