Define: $f: \mathbb{R}^{2} \rightarrow \mathbb{R}$ $f(x,y)$ for $(x,y) \ne (0,0)$ $$ \frac{x^{3}}{x^{2} + y^{2}} $$ and $f(x,y) = 0 \text{ for } (x,y) = (0,0)$
Question is the function differentiable in (0,0)?
So the definition my teacher gave me for differentiability is the following: $f$ is differentiable in $a \in E$ with derivative $A \in \operatorname{Hom}\left(\mathbb{R}^{n}, \mathbb{R}^{m}\right)$ if the following holds: $$ r(h):=f(a+h)-f(a)-A h $$
$$ \lim _{h \rightarrow 0} \frac{r(h)}{||h||}=0 $$
I have no idea what this definition means, especially why a matrix is in the definition??
Another definition I found on the internet is: $f$ is differentiable at $(0,0)$ if and only if: $$ \lim_{(x,y) \rightarrow (0,0} \frac{f(x,y)-f(0,0)-f_{x}(0,0)x-f_{y}(0,0)y}{\sqrt{x^{2}+y^{2}}} = 0 $$ So using this definition: my attempt is: $$ \lim_{(x,y) \rightarrow (0,0)} \frac{x^{3}}{x^{2}+y^{2}\sqrt{x^{2}+y^{2}}} = 0 $$ this limit exists because, $$ \frac{x^{3}}{x^{2}+y^{2}\sqrt{x^{2}+y^{2}}} \leq \frac{x^{3}}{x^{2}} = x < \delta = \epsilon $$ And this limit is $0$.
