I was studying a GRE-like test and came across the following task:
Suppose $f(x, y)$ is defined everywhere at $\mathbb{R}^n$ and $$\forall \ x, y \ \rightarrow \lim_{a \to 0} \frac{f(ax, ay)}{a} = x + y$$ then
A. $f$ is continuous at $(0, 0)$
B. $f$ is differentiable at $(0, 0)$
C. There are partial derivatives $\frac{\partial f(x, y)}{\partial x}$ and $\frac{\partial f(x, y)}{\partial y}$ at every points of some neighbourhood of $(0, 0)$
D. The function $g(x) = f(x, 0)$ is differentiable at $x = 0$ and $g'(0) = 0$
E. A-D are false
Honestly, I don't understand what that condition means. Does it mean that $f(x, y) = g(x+y)$, where $g(x) \sim x, x\to 0\ $?
I understand that B is false because otherwise A and C will be also true (I realize that it's not a rigorous proof, but only considerations)
Could you please give me any hints for this task? Thank you a lot in advance!
(A textbook says that E is true)
Consider $$f(x,y)=\begin{cases}x+y+\sin\left(\frac{y^4}{x}\right), &x\neq 0\\ y, &x=0\end{cases}.$$ Observe that for each $x\neq 0$, we have $$\lim_{a\to 0}\frac{f(ax,ay)}{a}=x+y+\lim_{a\to 0}\frac{\sin\left(\frac{a^3y^4}{x}\right)}{a}=x+y$$ For $x=0$, we have $$f(ax,ay)=ay$$ Thus clearly $f(x,y)$ satisfies the condition given. However it's easy to see that conditions A-D fail to satisfy:
A: Consider the curve $y^4=x$, and approach $(0,0)$ along this curve, we obtain the directional limit $$\lim_{y\to 0}y^4+y+\sin\left(\frac{y^4}{y^4}\right)=\sin(1)\neq 0.$$ Thus $f$ is not continuous at $(0,0)$.
Similarly you can prove that B-D also fail.