<s>[INST] The limit of functions of two variables 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?.
 [/INST] 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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