I need help finding if the limit of $g(x,y)$ at $(0,0)$ and at $(2,0)$ to see if it is continuous on these points.
I get confused, because normally these piecewise functions are defined such that $(x,y)$ different than some $(a,b)$ but here I have no idea how to do it because I can approach $(0,0)$ or $(2,0)$ from both pieces of the function...
And if I wanted to see the differentiability as well, what should I do?
I would really appreciate if you could explain this to me. Thank you !
On
For $\vert y \vert \le 1$ and $y \neq 0$, you have
$$\vert g(x,y) \vert \le \vert x y \vert + \vert x \vert \vert y \vert^3 \le 2 \vert x \vert \vert y \vert.$$
As the right hand side of this inequality goes to zero as $y \to 0$ and $g(x,0)=0$ for all $x$, we get that $g$ is continuous everywhere.
$g(x,y)$ is continuous at $(x_0,y_0)$ if $\forall \epsilon > 0$, $\exists \delta > 0$ $|| (x,y) - (x_0,y_0)|| < \delta \Rightarrow || g(x,y) - g(x_0,y_0)|| < \epsilon$.
Continuity at $(0,0)$: Let $\epsilon > 0$. Let $\delta = \min(\frac{\sqrt{\epsilon}}{2}, \frac{1}{2})$. \begin{align*} || (x,y) - (0,0)|| < \delta &\Rightarrow |x| < \delta \land |y| < \delta \\ \Rightarrow || g(x,y) - g(0,0)|| &= |xy + xy^3 \sin(\frac{x}{y})| \\ &\leq |xy| + |xy^3| < \delta^2 + \delta^4 \leq \epsilon \: (\text{check}!) \end{align*} Hence g is continuous at $(0,0)$. You could check for $(2,0)$.
$g(x,y)$ is differentiable at $(x_0,y_0)$ if $\exists (\alpha, \beta)$ s.t. $\epsilon(h_1, h_2) := \frac{g(x_0 + h_1, x_0 + h_2) - g(x_0,y_0) - (\alpha, \beta) \odot (h_1, h_2)}{||(h_1,h_2)||} \rightarrow 0$ as $(h_1, h_2) \rightarrow 0$. (where $\odot$ means dot product.)
You could check differentiability using this definition.
If you haven't seen these definitions before, then what you want to search for is 'continuity' and 'differentiability' of "multi-variable functions". Also there is another definition of continuity using sequences instead of $\epsilon-\delta$, you could again search for that (on google, or any calculus book on vector valued functions).
Edit (after seeing comment):
Intuitive understanding of continuity:
Continuity of $\mathbf{f: \mathbb{R} \rightarrow \mathbb{R}}$ at $\mathbf{x_0 \in \mathbb{R}}$.
Visualize $x_0$ on the real number line. The definition of continuity would mean "if you approach $x_0$ from any side, then it's corresponding value of $f(x)$ must approach $f(x_0)$. Note that since x is a real number, you can approach it from two sides - left and right leading to the definition of left hand limits and right hand limits etc.
Continuity of $\mathbf{f: \mathbb{R}^2 \rightarrow \mathbb{R}}$ at $\mathbf{(x_0,y_0) \in \mathbb{R}^2}$.
Visualize $(x_0, y_0)$ on the 2D plane (x-y plane). The definition of continuity would mean "if you approach $(x_0, y_0)$ from any side, then it's corresponding value of $f(x,y)$ must approach $f(x_0, y_0)$. Now however as $(x_0,y_0)$ is a point on the 2D plane, you can approach it from infinitely different directions. For example, you could approach this point via a line segment ending at $(x_0,y_0)$ with a slope of $k$ for any value $k$. (This actually leads to the definition of the $k$^th directional derivative). Alternatively you could approach it along any path which finally terminates at $(x_0,y_0)$ like a spiral etc.
If this is a bit hard to understand, let me know, I'll include some figures.