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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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Try along $x=y^2$. –  AJ Stas Aug 17 at 13:39

7 Answers 7

up vote 8 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

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

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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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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

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come along any $x=my^2$ then $\displaystyle \lim\limits_{(x,y)\to(0,0)}\frac{xy^2}{x^2+y^4}=\frac{1}{m^2+1}$, so for different $m$ you'll get different limits, so limit does not exists!

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There is no problem with polar coordinates, as long as the transformation is consistent. All of the mentioned substitution are path to reach $(x,y) \Rightarrow (0,0)$ i.e. simultaneous limits.

You can use polar coordinates: \begin{align*} &x=r(1+\cos(\theta)),y=r\sin(\theta) \\ &\Rightarrow lim_{\theta \rightarrow \pi} \frac{r^3(1+\cos(\theta))sin^2(\theta)}{r^2\{(1+\cos(\theta))^2+\sin^2(\theta)\}}=0 \end{align*}

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Could you elaborate on this method? –  Alizter Aug 18 at 17:32
its a simple substitution: The circle passing through origin i.e. passing the point required for the limiting case. –  Shark Aug 22 at 6:41

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