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I was trying to find if the the following limit exists: $$ \lim_{(x,y)\to (0,0)} \frac{x^3y^2}{x^6+y^4} $$ From this topic I tried to switch to polar coordinates: $$ \frac{x^3y^2}{x^6+y^4}=\frac{r^3\cos^3\theta\cdot r^2\sin^2\theta}{r^6\cos^6\theta+r^4\sin^4\theta}=\frac{r\cos^3\theta\sin^2\theta}{r^2\cos^6\theta+\sin^4\theta}\to 0 $$ But apparently this limit does not exists (proved it in another way). Why this method worked for the limit in that topic and here it didn't work?

vesii
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  • A comment on the linked answer using polar coordinates explains it is not a valid proof. It's an incorrect technique which in that case happened to give the correct answer. – aschepler Jun 13 '20 at 16:22

2 Answers2

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Consider the path along the x-axis, y-axis and $y=x^{\frac{3}{2}}$.

Along the x-axis: $$\lim_{(x,0) \to (0,0)} \frac{x^3 \cdot 0}{x^6 + 0}=0 $$

Along the y-axis:$$\lim_{(0,x) \to (0,0)} \frac{0 \cdot y^2}{0 + y^4}=0 $$

Along the curve $y=x^{\frac{3}{2}}$: $$ \lim_{(x,x^{\frac{3}{2}}) \to (0,0)} \frac{x^3{(x^{\frac{3}{2}})}^2}{x^6 + {(x^{\frac{3}{2}})}^4 }=\frac{x^6}{2x^6}=\frac{1}{2}$$

Therefore, the limit does not exist.

Ty.
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  • That's how I proved it does not exists. I know it does not exists. My question is why the converting to polar coordinates didn't work like it did in that other topic? – vesii Jun 13 '20 at 16:06
  • I believe that you're not supposed to treat $\theta$ as a constant. Are you supposed to use polar coordinates for this problem? Using the xy coordinate system is faster, easier, and more reliable. – Ty. Jun 13 '20 at 16:09
  • Just trying to understand why it didn't work and why it worked for them. They did treat it as a constant, then why their answer is correct? – vesii Jun 13 '20 at 16:13
  • @vesii The answer is correct, but the proof is not. Like the classic "to simplify $16/64$, cancel the sixes". – aschepler Jun 13 '20 at 16:26
  • The method does not work in this case, because at some point you conclude that the second term in the denominator is the dominant term. Taking the limit of r to zero then yields zero. However, you can take a different path to the origin in such a way that the first term in the denominator is dominant. This leads to a different result. And therefore we must conclude that the limit does not exist. – M. Wind Jun 13 '20 at 19:50
  • @aschepler Why is this proof incorrect? – Mark Viola Jun 14 '20 at 04:03
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The conversion to polar coordinates has not failed. Note what happens when $r=\tan^2(\theta)\sec(\theta)$, $|\theta| > \pi/2$. Then,

$$\frac{r\cos^3(\theta)\sin^2(\theta)}{r^2\cos^6(\theta)+\sin^4(\theta)}=\frac{\sin^4(\theta)}{2\sin^4(\theta)}=\frac12$$

And of course, if $\theta =0$ or $\theta =\pi/2$, then

$$\frac{r\cos^3(\theta)\sin^2(\theta)}{r^2\cos^6(\theta)+\sin^4(\theta)}=0$$

EXPLANATION: So what is happening here?

Note that $x^6+y^4\ge 2\sqrt{x^6y^4}=2|x|^3y^2$, with equality for $y^2=|x|^3$. Therefore, we have

$$\left|\frac{x^3y^2}{x^6+y^4}\right|\le \frac12$$

with equality when $y^2=|x|^3$.

If we transform to polar coordinates, this analysis is identical to

$$\left|\frac{r\cos^3(\theta)\sin^2(\theta)}{r^2\cos^6(\theta)+\sin^4(\theta)}\right|\le \left|\frac{r\cos^3(\theta)\sin^2(\theta)}{2\sqrt{r^2\cos^6(\theta)\sin^4(\theta)}}\right|=\frac12$$

with equality when $r=\frac{\sin^2(\theta)}{|\cos^3(\theta)|}=\tan^2(\theta)|\sec(\theta)|$.

Mark Viola
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  • @vesii Please let me know how I can improve my answer. I really want to give you the best answer I can. If it was not useful, then I can simply delete this answer if you prefer. – Mark Viola Aug 18 '20 at 02:23