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How to prove that $\lim_{(x,y) \to (0,0)} \dfrac{x^3y}{x^4+y^2} = 0?$

First I tried to contradict by using $y = mx$ , but I found that the limit exists.

Secondly I tried to use polar coordinates, $x = \cos\theta $ and $y = \sin\theta$,

And failed .. How would you prove this limit equals $0$?

Gigili
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Billie
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5 Answers5

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Note that, when $x$ and $y$ are both less than $1$, we have

$\displaystyle \left|\frac{x^2y}{x^4+y^2}\right| < \frac{1}{2}$

multiplying by $|x|$ we get

$\displaystyle \left|\frac{x^3y}{x^4+y^2}\right| < \frac{1}{2}|x|$

Result follows from squeeze theorem

dani_s
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  • You need $\le$, not $<$ (as shown by taking $y=x$). Also, you should explain how you derive the estimate. How is it relevant that $|x|<1$ and $|y|<1$ (I assume that you meant absolute values here)? – Hans Lundmark Sep 10 '18 at 01:12
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Cauchy inequality: $$x^4+y^2\ge 2 x^2 |y|$$

Thus $$ \left|\frac{x^3y}{x^2+y^4}\right|\le \frac{|x|}{2}. $$ But the RHS tends to $0$ as $(x,y)\to (0,0)$, and hence $$ \lim_{(x,y)\to(0,0)}\frac{x^3y}{x^2+y^4}=0. $$

Yiorgos S. Smyrlis
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Observe that $x^4 + y^2 \geq |x^2y|$ (for instance, because $x^4+y^2+2x^2y = (x^2+y)^2\geq0$ and $x^4+y^2-2x^2y = (x^2-y)^2 \geq0$). Hence $\displaystyle \left|\frac{x^2y}{x^4+y^2}\right| \leq 1$ when $(x,y)\neq (0,0)$ and thus $$\lim_{(x,y)\rightarrow (0,0)} \left|\frac{x^3y}{x^4+y^2}\right| \leq \lim_{(x,y)\rightarrow(0,0)} |x| = 0,$$ so the limit is 0 by the squeeze theorem.

universalset
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Let $y=kx^2$, then the expression is $\frac{kx}{1+k^2}$.
$|\frac{k}{1+k^2}|\leq 1/2$ so the expression is bounded by $|x/2|$. The path along $x=0$ can be done separately.

Empy2
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With polar coordinates we're cool, too:

$$x=r\cos\theta\;\;,\;\;y=r\sin\theta$$

$$\frac{x^3y}{x^4+y^2}=\frac{r^4\cos^3\theta\sin\theta}{r^4\cos^4\theta+r^2\sin^2\theta}=\frac{r^2\cos^3\theta\sin\theta}{r^2\cos^4\theta+\sin^2\theta}\xrightarrow[r\to 0]{}\frac0{0+\sin^2\theta}=0$$

But what if $\;\sin^2\theta=0\;$ ? Well, then also $\;\sin\theta=0\;$ and the first expression above's already zero from the beginning...

DonAntonio
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    -1. This isn't a correct argument, since it would also “prove” that $x^2 y/(x^4+y^2) \to 0$, which is false (since $x^2 y/(x^4+y^2) = 1/2$ for $y=x^2 \neq 0$). The problem is that you can't treat $\theta$ as a constant when letting $r \to 0$. – Hans Lundmark Sep 10 '18 at 01:09