# Multivariable Limit - Converting to Polar Coordinates

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.

• Using polar coordinates is a trap. Just find the limits along the line $y=x$ and $y^2=x$ – Bor Kari Mar 16 at 0:59

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