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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!

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  • $\begingroup$ 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) $\endgroup$ – Sonal_sqrt Aug 26 '19 at 7:17
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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 $r \to 0$).

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