Why cannot we say The limit $\lim _{(x, y) \rightarrow(0,0)} \frac{x y^{2}}{x^{2}+y^{4}}=0$

Question

I used Polar coordinates to evaluate $$L=\lim _{(x, y) \rightarrow(0,0)} \frac{x y^{2}}{x^{2}+y^{4}}$$

We have $x=r\cos \theta$ and $y=r\sin \theta$ and as $(x,y)\to (0,0)$ we have $r \to 0$ So we get $$L=\lim_{r \to 0}\frac{r\cos \theta \sin^2\theta}{\cos^2\theta+r^2\sin^4\theta}=0$$

So irrespective of any path, the limit is zero. I know that this is wrong since the limit actually does not exist, But i have seen some threads in this forum. Here is what i understood.

Even though $r \to 0$, the quantity $$f(r, \theta)=\frac{ \cos \theta \sin ^{2} \theta}{\cos ^{2} \theta+r^{2} \sin ^{4} \theta}$$ might be infinity for some of the paths other than the straight line. For example if we choose the path $x=y^2$, then when $r \to 0$ we get $\theta \to \frac{\pi}{2}$. So the quantity $\frac{\cos \theta \sin ^{2} \theta}{\cos ^{2} \theta+r^{2} \sin ^{4} \theta}$ blows up to infinity. So the limit Does not exist. Is this the correct analysis? Any inputs please add.

Answer 1

Your analysis looks to be mostly correct (although it could be clearer and more precise) that the limit doesn't exist, but there isn't a need to introduce $\sin$ and $\cos$. Following the path $x=y^2$ means that you need to compute the limit: $$ \lim_{y\rightarrow 0}\frac{y^2y^2}{(y^2)^2+y^4}=\lim_{y\rightarrow 0}\frac{2y^4}{y^4}=2. $$

Answer 2

This claim:

$$L=\lim_{r \to 0}\frac{r\cos \theta \sin^2\theta}{\cos^2\theta+r^2\sin^4\theta}=0$$

is not justified. Perhaps we follow a path where $\theta$ is a function of $r$, and it's just not clear that the limit will be $0$. In fact as you found, for some paths the limit is not $0$.