1. ## Quick question

Let $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ be a function such that along every line $y=Ax$ and the line $x=0$, the function is continuous at $(0,0)$. By this I mean if we let $L_{\theta}$ be one of the lines in the x-y plane given by rotating an angle $\theta$ anti-clockwise from the x-axis, and let $\{ a_n \}$ be a sequence converging to $(0,0)$ s.t. $a_n\in L_{\theta}$ for all $n$, then $f(a_n) \rightarrow f(0,0)$ as $n \rightarrow \infty$.

Is it true then that $f$ is continuous at $(0,0)$? So if we take an arbitrary sequence converging to $(0,0)$, say $\{ b_n \}$, then does $f(b_n) \rightarrow f(0,0)$? I was thinking that the each element $b_n$ lies on some line $L_{\theta_n}$ so that there are countably many lines that the sequence lies on, but I'm not sure if that means anything. Is the statement even true?

Thanks for any help, just something that was bothering me

2. ## Re: Quick question

Hint: suppose the slope of $f$ restricted to the line $L_\theta$ is $m_\theta$, and suppose there's a sequence $\theta_i$ such that $m_{\theta_i}\to\infty$.

3. ## Re: Quick question

If I understood this, take $f(x,y)=\frac{xy^2}{x^2+y^4}$. It has limit zero on every line through the origin, but if you approach zero through the parabola $x=y^2$ the limit is $\frac{1}{2}$

4. ## Re: Quick question

Originally Posted by Jose27
but if you approach zero through the parabola $x^2=y$ the limit is $\frac{1}{2}$
Did you mean $x=y^2$ ?