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Show that the limit of the function, $f(x,y)=\frac{xy^2}{x^2+y^4}$, does not exist when $(x,y) \to (0,0)$.

I had attempted to prove this by approaching $(0, 0)$ from $y = mx$, assuming $m = -1$ and $m = 1$. The result was $f(y, -y) = \frac{y}{1+y^2}$ and $f(y, y) = \frac{y}{1+y^2}$ as the limits which are obviously different. Essentially, I was just wondering what is the correct working out for a solution to this question.

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  • $\begingroup$ Welcome to Math.SE! Can you show what you have tried yourself? This helps other people give better answers and sets your question apart from the multitude of homework questions that appear on this site where no effort has been put in. $\endgroup$ – Hrodelbert Aug 26 '15 at 10:52
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    $\begingroup$ Sorry! will do. $\endgroup$ – Tom Stewart Aug 26 '15 at 10:59
  • $\begingroup$ Why are the limits different? They are both zero. Can you think out a path that has a nonzero (and finite) limit? $\endgroup$ – uranix Aug 26 '15 at 13:50
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I assume that you would like to see either that the function is not continuous at $(0,0)$ or that the function is not defined at $(0,0)$.

That the function is not defined at $(0,0)$ holds because we do not have division by zero in the real field.

Let $f: (x,y) \mapsto xy^{2}/(x^{2} + y^{4})$ on $\mathbb{R}^{2}\setminus \{ (0,0) \}$ and let $f(0,0) := 0$. We claim that the function is not continuous at $(0,0)$. Note that along every straight line through the origin it holds that $f(x,y) \to 0$ as $(x,y) \to (0,0)$, i.e. for every $m \in \mathbb{R}$, along set $S_{m} := \{ (x,y) \mid y = mx \}$ we have $f(x,y) \to 0$ as $(x,y) \to (0,0)$, i.e. $f\mid_{S}(x,y) \to 0$ as $(x,y) \to (0,0)$. Thus $f(0,0)$ is the limit of $f$ at $(0,0)$ along every straight line through the origin. However, since the function $f$ restricted on some curve, say on the parabola $T := \{(x,y)\mid y^{2} = x \}$, takes the value 1/2, and since for every neighborhood $U$ of $(0,0)$ we have $U \cap T \neq \varnothing$, it follows that $f\mid_{T}(x,y) \to 1/2 \neq f(0,0)$ as $(x,y) \to (0,0)$, whence $f$ is not continuous at $(0,0)$.

I hope the above argument is helpful.

Please note that if your original question is to ask if the limit of $f$ at $(0,0)$ exists then the above argument also covers your question, for it shows that the limit of $f$ at $(0,0)$ may or may not exist, dependent on the choice of paths along which we are talking about the limit.

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  • $\begingroup$ Thank you. If you do not mind, in fact every user has the right to up-vote or down-vote any question or answer out of other people. :P $\endgroup$ – Megadeth Aug 26 '15 at 12:28
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$$\lim\limits_{(x,y)\to (0,0)}\frac{xy^2}{x^2+y^4}$$ Using polar coordinates, we have $$\lim\limits_{r\to 0^+}\frac{r^3\cos\phi\sin^2\phi}{r^2\cos^2\phi+r^4\sin^4\phi}$$ $$=\lim\limits_{r\to 0^+}\frac{r\cos\phi\sin^2\phi}{\cos^2\phi+r^2\sin^4\phi}$$ Now lets try to find bounds that are independent of $\phi$. Since $$\left|\cos\phi\sin^2\phi\right|\leq 1$$ We have $$\frac{r\left|\cos\phi\sin^2\phi\right|}{\left|\cos^2\phi+r^2\sin^4\phi\right|}\leq \frac{r}{\left|\cos^2\phi+r^2\sin^4\phi\right|}$$ Notice that the right hand side cannot be bounded in terms that are independent of $\phi$. Therefore $$\lim\limits_{(x,y)\to (0,0)}\frac{xy^2}{x^2+y^4}=\mbox{non existent}$$

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