Let
$$f(x,y) =
\begin{cases}
0, & |y| > x^{2}\ \text{or}\ y = 0 \\
1, & \text{otherwise}
\end{cases}
$$
Show $\mathbf{f}$ is continuous at $(0,0)$ on every line through the origin but not is not continuous at $(0,0)$.
Attempt:
Observe that a line is described as $y = cx$. Also for $|x| < 1$, $|cx| > x^{2}$ is always true. As a result of this we will always be in the case where $|y| > x^{2}$.
Choose $\delta = \frac{\epsilon}{|c|}$ for any $c \in \mathbb{R}$. Observe that $|x-0|, |y-0| \leq \|(x,y) – (0,0)\| < \delta$.
With this in place:
$$\|f(x,y) – f(0,0)\| = |cx – 0| = |c||x| < |c|\delta = |c| \frac{\epsilon}{|c|} = \epsilon$$
Therefore along any line through the origin $\mathbf{f}$ is continuous.
The reason $\mathbf{f}$ is not continuous at $(0,0)$ is because in the case $|y| < x^{2}$, $f(x,y) = 1$. So the limits will never be equal.
End of proof
Issue: The concern I have with my proof is how I showed continuity in the linear case. I don't know if I expressed the right idea. I visually see what is happening. In words: I know any linear path I take towards the origin will be the within $|y| > x^{2}$ case. As such the function should always equal $0$.
But when drawing pictures and looking at the domain, the $\delta$ that I choose will have to depend on the level of the slope of my line, i.e the $c$ value.
So my main issue is that the function is defined as the original one at the top, but I used a different function in showing the epsilon statement. I'm trying to liken the idea to say a situation where for example we:
$$f(x,y) = \frac{x^{2} + y^{2}}{x}$$
If I take the limit $(x,y) \to (0,0)$ along the path $y = cx$, I would put that $y$ value in my expression and take the limit or in particular to this situation I could find a $\delta$ to show it is continuous along lines only. I'm not sure I'm communicating the same idea correctly in the above case. Perhaps I'm over complicating it. (I'm just using this as an example for what I'm doing above) Thoughts?
Best Answer
There are several problems here.
Firstly it is not true that $\vert cx\vert > x^2$ for $\vert x\vert < 1$. Take $x$ close to $1$ and $c$ close to zero and the inequality won't hold.
Secondly, you wrote : $$ \vert f(x,y) - f(0,0) \vert = \vert cx - 0\vert.$$
This equality is incorrect since $f(x,y)$ is either $1$ or $0$. Where does this $cx$ come from ?
Anyway these problems can be fixed.
We want to prove that $$ \lim_{(x,y =cx) \to (0,0)} f(x,y) = 0$$ for any value of $c$. Clearly this is true for $c = 0$.
Suppose that $c\neq 0.$ Notice that $ \vert cx\vert > \vert x\vert^2 \iff \vert x\vert ^2- \vert c \vert \vert x \vert < 0 \iff \vert x\vert \in (0, \vert c \vert)$.
Let $\epsilon >0$.
$$\Vert (x,cx)\Vert < \delta := \vert c \vert \implies \vert x \vert < \vert c \vert \implies f(x,cx) = 0 \implies \Vert f(x,cx)- f(0,0)\Vert = 0 < \epsilon.$$
Hence the limit is $0$ along the lines $y = cx$.
The fact that $\delta$ depends on the slope is not a problem. In fact we expect $\delta$ to depend on $c$ since we're taking a limit along a path which is dependent on $c$.