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The limit of $f(x,y)=\frac{xy^2}{x^2+y^4}$ as $(x,y) \longrightarrow(0,0)$ is doesn't exist, because if we take two paths:

1) Along the path, $x=0$, $y\longrightarrow0$: $$\lim_{(x,y)\longrightarrow(0,0)}\frac{xy^2}{x^2+y^4}=0$$ 2) Along the path, $x=y^2$, $y\longrightarrow0$: $$\lim_{(x,y)\longrightarrow(0,0)}\frac{xy^2}{x^2+y^4}=\lim_{y\longrightarrow0}\frac{y^2y^2}{(y^2)^2+y^4}=\frac{1}{2}$$

But if we use polar coordinate methods of evaluating limits of functions of two variable, we get the limit of the above function becomes zero, i.e., $x=r\cos(\theta),y=r\sin(\theta)$, we know that $x^2+y^2=r^2$ and this indicates that $r\longrightarrow0$ as $(x,y)\longrightarrow(0,0)$, therefore $$\lim_{(x,y)\longrightarrow(0,0)}\frac{xy^2}{x^2+y^4}=\lim_{r\longrightarrow0}\frac{r^3\cos(\theta)\sin^2(\theta)}{r^2\cos^2(\theta)+r^4\sin^4(\theta)}$$ $$=\lim_{r\longrightarrow0}\frac{r\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)+r^2\sin^4(\theta)}=0$$ Please someone help me on this contradictory, why this is so happened?.

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  • $\begingroup$ When calculating with $r \to 0$, you assume that $\theta$ is constant. So you follow paths on a straight line to the origin. But for calculating the limit, you have to calculate it for $\textit{every}$ path to $(0,0)$. $\endgroup$ – Jan Apr 1 at 9:03
  • $\begingroup$ Another viewpoint: the convergence when $r\to 0$ must be uniform in $\theta$. $\endgroup$ – Martín-Blas Pérez Pinilla Apr 1 at 15:56
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When you fixing $\theta$ and take limit as $ r \to 0$ you are taking limits along straight lines. It is not enough to take limits along these lines. When you take limit along the parabola $x=y^{2}$ we get a different value for the limit so the limit of the function does not exist.

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You need to be careful and consider all paths as well, not only paths with constant $\theta$.

$$=\lim_{r\longrightarrow0}\frac{r\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)+r^2\sin^4(\theta)}$$

$$=\lim_{r\longrightarrow0}\left(r\color{blue}{\frac{\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)+r^2\sin^4(\theta)}}\right)$$

Your argument works if the expression in blue stays bounded (near the origin); for all $(r,\theta)$.


Inspired by your own path $x=y^2$, check what happens in polar coordinates ($r \ne 0$): $$x=y^2 \longrightarrow r\cos\theta=r^2\sin^2\theta \implies r=\frac{\cos\theta}{\sin^2\theta}$$ Along $r=\frac{\cos\theta}{\sin^2\theta}$ you can take $\theta\to\tfrac{\pi}{2}$ to get $r\to 0$, but what happens with your function along that path?

$$\lim_{r\to 0}\frac{r\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)+r^2\sin^4(\theta)}\stackrel{r=\frac{\cos\theta}{\sin^2\theta}}{\longrightarrow}\ldots$$

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I asked this question a couple of days ago. I want to answer my question myself, I'm not fully satisfied with your answers. Let me get started by stating the following Theorem with out proof.

Theorem 1. If the limits of $f(x,y)$ and $g(x,y)$ exists as $(x,y)\rightarrow(0,0)$, then the limit of the product $f(x,y)g(x,y)$ exists as $(x,y)\rightarrow(0.0)$ and is equal to the product of individual limits.

Theorem 2. If the limits of $f(x,y)$ exists and the limits of $g(x,y)$ doesn't exist other than $\pm \infty$, then the limit of the product $f(x,y)g(x,y)$ doesn't exist.

Theorem 1 states that to be the limit of the product exists the limits of the individual functions must exist.

Now let me apply the Theorem 1 and 2 to see the above example has no contradiction with the two methods, using paths and using polar coordinate method. As we have seen, the limit doesn't exist since we got two different limits for the same function along two different paths. Now let's check why the limit is doesn't exist when we use polar coordinate.

Let $x=r\cos(\theta), y=r\sin(\theta)$, we know that $x^2+y^2=r^2$, this indicates that $r\rightarrow 0$ as $(x,y)\rightarrow(0,0)$. Therefore $$\lim_{(x,y)\rightarrow(0,0)}\frac{xy^2}{x^2+y^4}=\lim_{r\rightarrow 0}\frac{r^3\cos(\theta)\sin^2(\theta)}{r^2\cos^2(\theta)+r^4\sin^4(\theta)}$$ $$=\lim_{r\rightarrow 0}\frac{r\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)+r^2\sin^4(\theta)}\ne\bigg(\lim_{r\rightarrow 0}r\bigg)\bigg(\lim_{r\rightarrow 0}\frac{\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)+r^2\sin^4(\theta)}\bigg)$$ As you see the right hand equation, which is the product of two limits $\lim_{r\rightarrow 0}r$ which exists, but the limit $\lim_{r\rightarrow 0}\frac{\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)+r^2\sin^4(\theta)}$ doesn't exist, that is why I put $\ne$ in between according to Theorem 1. Therefore in this case we can't apply Theorem 1 and by Theorem 2 the limit doesn't exist.

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