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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

Asaf Karagila
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tkhu
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2 Answers2

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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.

Najib Idrissi
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Another approach:

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

DonAntonio
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    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. – Steven Stadnicki Feb 13 '14 at 16:45
  • @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. – DonAntonio Feb 13 '14 at 16:48
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    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. – Steven Stadnicki Feb 13 '14 at 16:55
  • 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.. – DonAntonio Feb 13 '14 at 17:02
  • Here Clark shows how to do this without any handwaving. – Jyrki Lahtonen Sep 19 '15 at 23:37
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    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. – Jyrki Lahtonen Sep 20 '15 at 06:04