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

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3 Answers 3

up vote 5 down vote accepted

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.

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Great! many thanks. Regards. –  mathsalomon Jul 23 '12 at 11:45

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 $0$.

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Yes, I had not noticed. –  mathsalomon Jul 23 '12 at 11:46

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.

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