If $\displaystyle \ \ f(x,y) = \begin{cases} \frac{3x^2y}{x^2+y^2}, &(x,y) \neq (0,0) \\ a, &(x,y)=(0,0) \end{cases}$
find the value of $a$ such that $f$ is continuous at the origin.
I don't want an answer. This was just an example. (similar question was asked in my university last semester and there was a debate between professors)
Should the student prove that $\displaystyle \lim_{(x,y) \to (0,0)} f(x,y)$ exists?
Or is it enough for them to find a limit along one line through the origin and say that $f(0,0)$ should be equal to it? (because the limit on every path should exist and equal to that limit, otherwise the question becomes wrong)
Assume we can indeed find such $a$ that makes $f$ continuous, we will have
$$ a = f(0,0) = \lim_{(x,y) \to (0,0)} f(x,y) = \lim_{x \to 0} f(x,0) = 0 $$
so the only possible value for $a$ is zero. However, this doesn't prove that if we set $a = 0$, we get a continuous function $f$ (since we checked the existence of the limit and the continuity only along the $x$-axis). In my opinion this means that the student should prove now that this definition indeed makes $f$ continuous (for example, by noting that
$$ 0 \leq \left| \frac{x^2y}{x^2 + y^2} \right| = |y| \frac{x^2}{x^2+y^2} \leq |y| $$
and since $\lim_{(x,y) \to (0,0)} |y| = 0$, by sandwich we also have $\lim_{(x,y) \to (0,0)} f(x,y) = 0 = f(0,0)$).