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Given the function: $$f(x) = \begin{cases} \frac{x^2y}{x^4+y^2} &\mbox{if } (x,y) \ne (0,0) \\ 0 & \mbox{if } (x,y)=(0,0). \end{cases} $$ Show that function restricted to random line passing through point (0,0), is on this line continuous but the point (0,0) is not a point of continuity of this function.

To show that the function is not continuous in the point is simple (limit should be enough), but how to show that on the whole line?

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If you take a line passing through point $(0,0)$ that mean it exist $a\in \mathbb{R}$ such that $$ y=ax $$ So in such line $f(x,y)$ is equal to : $$ f(x,y)=\left\{\begin{array}{ccc} 0 &\textrm{ if }& x=0 \\ \frac{ax}{x^2+a^2} & \textrm{ if }& x\neq 0 \end{array}\right. $$ and it's clear that $\lim_{x\to 0} f(x,y)=0$ for all $a\in\mathbb{R}$, so the function is continuous in any line passing through the point $(0,0)$.

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You can also parametrize a line through $(0,0)$ as $(r\cos(\theta),r\sin(\theta))$ with $\theta$ fixed and $r$ the parameter. Then $$f(x,y)=\frac{r\cos^2(\theta)\sin(\theta)}{r^2\cos^4(\theta)+\sin^2(\theta)}$$

To show that $f$ is continuous through any line through the origin, show that $\lim_{r\rightarrow 0}f(x,y)=0$.

To show that $f$ is not continuous at the origin, find a $\delta>0$ so that no matter how small $r$ is, there is a $\theta$ so that $|f(r,\theta)|>\delta$. Try $\delta=0.25$ and $\theta$ just larger than $0$.

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