I need some insight into the “approach” that I used to solve this problem. Namely, I was asked to find if the following function is continuous on all $\mathbb{R}^2$:
$$ f(x, y) = \left\{ \begin{array}{ll} \frac{2 x^2}{x^2 + y^2} & \text{, if } (x, y) \neq (0, 0)\\ 0 & \text{, if } (x, y) = (0, 0) \end{array} \right. $$
I began by stating that the only point of concern is $(0, 0)$ and defined continuity as so:
A function $f(x, y)$ is continious at the point $(x_0, y_0)$ if and only if $f(x_0, y_0)$ exists, $\lim\nolimits_{(x, y) \to (x_0, y_0)} f(x, y)$ exists and $\lim\nolimits_{(x, y) \to (x_0, y_0)} f(x, y) = f(x_0, y_0)$.
Then I proceeded to show that $\lim\nolimits_{(x, y) \to (0, 0)} f(x, y) = \lim\nolimits_{(x, y) \to (0, 0)} \frac{2 x^2}{x^2 + y^2}$ does not exists by using limits along curves:
Consider the x-axis. Its parametric equation is $x = t, y = 0$, and $(x, y) \to (0, 0) \Rightarrow t \to 0$. This gives us the following limit: $$ \lim\limits_{(x, y) \to (0, 0)} \frac{2 x^2}{x^2 + y^2} = \lim\limits_{t \to 0} \frac{2 t^2}{t^2 + 0^2} = 2 $$
Next, consider the curve (line) $y = x$. Its parametric equation is $x = t, y = t$, and $(x, y) \to (0, 0) \Rightarrow t \to 0$. This gives us the following limit: $$ \lim\limits_{(x, y) \to (0, 0)} \frac{2 x^2}{x^2 + y^2} = \lim\limits_{t \to 0} \frac{2 t^2}{t^2 + t^2} = 1 $$
Since we got two different results as we approached $(0, 0)$ by two different curves, we can conclude that the general limit does not exist at that point. This means that $f(x, y)$ is not continuous at $(0, 0)$ and hence not continuous on $\mathbb{R}^2$.
Is everything I did here correct?
Thank you for your time.
