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I am new to this concept, but I do know that, using Cartesian coordinates, if the limit is different for 2 different "routes", then it does not exist.

I need to show that $$\lim_{(x,y)\to(0,0)}\frac{xy^2}{x^2+y^4}$$ DNE by converting to polar.

However, the result I've got is that the limit does exists, and it's equal to $0$. Here's what I did: $$ x=r\cos\theta\\ y=r\sin\theta\\ $$ $$ \lim_{(x,y)\to(0,0)}\frac{xy^2}{x^2+y^4}= \lim_{r\to0}\frac{r\cos\theta (r\sin\theta)^2}{(r\cos\theta)^2+(r\sin\theta)^4} =\lim_{r\to0}\frac{r^3\cos\theta \sin^2\theta}{r^2\cos^2\theta+r^4\sin^4\theta} =\lim_{r\to0}\frac{r^\require{cancel}\cancel{3}\cos\theta \sin^2\theta}{\require{cancel}\cancel{r^2}(\cos^2\theta+r^2\sin^4\theta)} =\lim_{r\to0}\frac{r\cos\theta \sin^2\theta}{\cos^2\theta+r^2\sin^4\theta}=\frac{0}{\cos^2\theta}=0 $$ Apparently no dependency on $\theta$? I mean, it is possible that $\cos^2\theta=0$. How do I continue from here? Am I missing something?

Thank you very much.

Robert Howard
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Netanel
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1 Answers1

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Your computation shows that, for example, if you enter the origin along $\theta=0$, your limit is zero. If you enter the origin along the curve $\cos\theta=r\sin^2\theta$ (which is tangent to the $Y$-axis as $\theta\to\pi/2$, your limit is $1/2$ (just replace $r\cos\theta$ by $r^2\sin^2\theta$ in your formulas before you cancel $r$).

GReyes
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  • Sorry, I didn't understand? Can you explain further? Why should I replace $r\cosθ$ by $r^2\sin^2θ$? – Netanel Mar 16 '19 at 17:32