[Math] continuity and differentiability of function of two variables

Question

Let $f(x,y)$ be
$$f(x,y): \begin{cases} x & \text{for } y = 0\\ x-y^3\sin\left(\frac{1}{y}\right)& \text{for } y \neq 0\end{cases} $$

then check continuity and differentiability at $(0,0)$.

Answer 1

$$ f(x,y) = x - y^3\sin(.) = x + o(x,y) $$ because $$\lim_{(x,y)\to (0,0)}\frac {y^3\sin(.)}{\sqrt{x^2 + y^2}} =0$$

• For continuity: $$\lim_{(0,0)} f(x,y) = \lim_{(0,0)}x + \lim_{(0,0)}o(x,y) =0 $$ then $f$ is continunous in $(0,0)$.
• For differentiability: Now identify with the definition of the derivative: $$ f(x,y) = f(0,0) + Df(0,0)\cdot(x,y) + o(x,y) $$ gives you the result: $f$ is diffrentiable in in $(0,0)$ and $Df(0,0)\cdot(x,y) = x$.