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

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

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 $$\theta =\dfrac {k\pi}2$$.

• This is what I got but what if $\theta = \pi / 2$ does it create problems? – DanieleMS Jul 19 at 11:26
• I don't think so. See my edit. – Chris Custer Jul 19 at 11:45
• So cool thanks I had not thought of that! – DanieleMS Jul 19 at 11:53
• 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. – David K Jul 19 at 17:10
• @DavidK good catch. I overlooked that. Plus I now see that I didn't need polar. – Chris Custer Jul 19 at 19:57

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 $$0$$.

• Nice answer! So simple! – quasi Jul 20 at 1:29

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 $$\lim_{(x,y)\to (0,0)}f(x,y)=0 \qquad\qquad\qquad\qquad\qquad\;\;\;$$

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)

• 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. – quasi Jul 19 at 13:20
• In any case, as has been shown in other answers, the limit is, in fact, equal to $0$. – quasi Jul 20 at 1:32