# Conditions to exploit Polar coordinates in limits.

Evaluate, $$\lim_{(x,y)\rightarrow(0,0)}f(x,y)=\lim_{(x,y)\rightarrow(0,0)}\dfrac{2x^2y}{x^4+y^2}$$

When I used polar coordinates with $$x=r\cos\theta, y=r\sin\theta$$,

$$\lim_{r\rightarrow0}\dfrac{r\cos\theta\sin2\theta}{r^2\cos^4\theta+\sin^2\theta}=0$$

But when I use path $$y=x^2$$,

$$\lim_{(x,y)\rightarrow(0,0)}\dfrac{2x^4}{2x^4}=1$$

Also from path $$x=0$$ or $$y=0$$ both gives, $$\lim_{(x,y)\rightarrow(0,0)}\dfrac{2x^2y}{x^4+y^2}=0$$

From path knowledge, I can say Limit does not exist.

Why this occurred that I got two different values of limits from Polar and the path makes me put a question that when to employ polar coordinates method to compute limits? When can I ascertain that it gives the correct value? Why is it giving out the value $$0$$ even when limit DNE?

Please help!

• polar coordinates will be useful if degree of polynomial in numerator is larger than denominator. In such a case limit usually exists like. It is not a rigorous rule just a common observation on my part. Also if degree in dnm is equal or greater than numerator limit almost always fails to exist.(again an observation not a theorem) – Sonal_sqrt Aug 26 '19 at 7:17

## 1 Answer

If in polar coordinates the function takes the form $$g(r) \, h(r,\theta)$$ where $$g(r) \to 0$$ as $$r \to 0^+$$ (standard single-variable limit) and the function $$h$$ is bounded for all $$\theta$$ and all $$r$$ in some region $$0 < r < R$$, then you can draw the conclusion that the two-variable limit is zero.

But that's not what you have in your case. Sure, you get a factor $$r$$ which tends to zero, but the remaining expression isn't bounded (as your other argument with $$y=x^2$$ shows; no matter how small $$r$$ is, you can find a $$\theta$$ such that the whole expression equals $$1$$, i.e., that other part standing together with $$r$$ is equal to $$1/r$$, which is unbounded as $$r \to 0$$).