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I was asked to find $$\lim_{(x,y)\rightarrow(0,0)}\frac{x^3y^2}{x^4+y^6}$$

Observe that setting y=mx results in $$\lim_{(x,mx)\rightarrow(0,0)}\frac{x^3(mx)^2}{x^4+(mx)^6} = 0$$ The textbook solution then proved that the limit is 0 using the squeeze theorem.

However, I tried to set y=x^(4/6) and I got: $$\lim_{(x,y)\rightarrow(0,0)}\frac{x^3(x^{\frac{4}{6}})^2}{x^4+(x^{\frac{4}{6}})^6} = \lim_{x\rightarrow0}\frac{x^4}{x^4+x^4} = \frac{1}{2}$$

So I concluded that the limit does not exist. I am not convinced that my solution is correct, I would really appreciate to know the reason why I am wrong.

Thank you

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  • $\begingroup$ Please don't use display math (double dollar signs) in titles, it makes the frontpage very unsightly. $\endgroup$ – Najib Idrissi Feb 13 '14 at 15:41
  • $\begingroup$ Yup, I'll be more careful from now on. $\endgroup$ – tkhu Feb 13 '14 at 15:43
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    $\begingroup$ Related. I think the argument I proffered there is somehow more convincing than an attempt to use the polar coordinates. $\endgroup$ – Jyrki Lahtonen Sep 19 '15 at 23:17
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Be careful in your computations... $(x^{4/6})^2 = x^{4/3}$ and then $x^3 x^{4/3} = x^{13/3}$. The end result of the expression is $$\frac{x^{13/3}}{2x^4} = \frac{1}{2}x^{1/3}$$ which does converge to zero.

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Another approach:

$$\frac{x^3y^2}{x^4+y^6}\stackrel{\text{Polar Coord.}}=\frac{r^5\cos^3\theta\sin^2\theta}{r^4(\cos^4\theta+r^2\sin^6\theta)}=r\frac{\cos^3\theta\sin^2\theta}{\cos^4\theta+r^2\sin^6\theta}\xrightarrow[r\to0]{}0$$

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    $\begingroup$ What happens when the origin is approached along a path where $\cos\theta\to0$? Then the denominator here goes to zero, potentially faster than the numerator, and it's not immediatly clear at all that the limit must be zero. $\endgroup$ – Steven Stadnicki Feb 13 '14 at 16:45
  • $\begingroup$ @StevenStadnicki, in that case the numerator approaches zero whereas the denominator approaches $\;0+r^2\cdot 1\;$ , and in the overall the limit is zero... The numerator does not approach zero in this case. $\endgroup$ – DonAntonio Feb 13 '14 at 16:48
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    $\begingroup$ Right, but we're also taking the limit as $r\to0$, remember, so the limit of $0+r^2\cdot1$ is also zero. I just don't see how this gets you away from a $\frac00$ term at all. $\endgroup$ – Steven Stadnicki Feb 13 '14 at 16:55
  • $\begingroup$ Either the path where $\;\cos\theta\to 0\;$ is limited, and then we go by $\;r\;$, or else we stick on that path (say, $\;\theta=\frac\pi4\;$) , and then from the start we have zero.. $\endgroup$ – DonAntonio Feb 13 '14 at 17:02
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    $\begingroup$ I ended up downvoting this. Your handwavy argument seems to suggest that $f(x,y)/r$ would be bounded near the origin. But, when $x=t^3, y=t^2$ we see that $f(x,y)=t/2$. In other words $f(x,y)$ can be as small as a constant times $\sqrt{r}$. Addressing @Steven's critic by talking about paths is unconvincing. One might think that you claim $f(x,y)/r$ to be bounded near the origin, which is false. $\endgroup$ – Jyrki Lahtonen Sep 20 '15 at 6:04

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