Prove that $\lim\limits_{(x,y) \to (0,0)} \frac{{x{y^2}}}{{{x^2} + {y^4}}} = 0$ [duplicate]

$$\lim_{(x,y) \to (0,0)} \frac{{x{y^2}}}{{{x^2} + {y^4}}} = 0$$ Please, Anyone could suggest me some way for this?. Thanks.

If indeed the limit was zero then every way we approach $(0,0)$ the limit would have to be $0$.

However, if we take the path $x=y^2$ we have:

$$\lim_{x\to 0}\frac{x^2}{x^2+x^2}=\frac12\neq 0$$

So the limit cannot be zero. Maybe it could be something else, but then it would have to be $\frac12$. Take $y=0$, we have:

$$\lim_{x\to 0}\frac0{x^2}=0\neq\frac12$$

Therefore the limit does not exist.

If $y=x$ the limit is $0$, but if $x=y^2$ the limit is $\frac{1}{2}$, then the limit don't exist.

Let $x=y^2$ therefore we have $$\lim\limits_{(x,y)\to(0,0)}\frac{xy^2}{x^2+y^4}=\lim\limits_{y\to0}\frac{y^4}{2y^4}=\frac{1}{2} \ \ \ (1)$$ and if we consider $x=y$ then $$\lim\limits_{(x,y)\to(0,0)}\frac{xy^2}{x^2+y^4}=0 \ \ \ \ (2)$$ so that from (1) and (2) we see that limit don't exits

This is not true. Consider sequence $(x_n,y_n)=(n^{-2},n^{-1})$ then you get $$ \lim\limits_{n\to\infty}(x_n,y_n)=0\\ \lim\limits_{n\to\infty}\frac{x_n y_n^2}{x_n^{2}+y_n^{4}}=1 $$ If you consider another sequence $(x_n,y_n)=(n^{-1},n^{-1})$ then you get $$ \lim\limits_{n\to\infty}(x_n,y_n)=0\\ \lim\limits_{n\to\infty}\frac{x_n y_n^2}{x_n^{2}+y_n^{4}}=+\infty $$ So we conclude that limit $$ \lim\limits_{(x,y)\to(0,0)}\frac{xy^2}{x^2+y^4} $$ even doesn't exist, not to mention it is equal to $0$.

Let $f(x,y):=\frac{xy^2}{x^2+y^4}$. We have $f(x^2,x)=1/2$ and $f(0,y)=0$, which proves that the limit doesn't exist.