Given a function$$f \in C^\infty:R^n\to R \text{ such that } f(x) = \frac{||x||^2}{2} + o(||x||^2) \text{ and }(n \geq 2)$$
The lecture I'm taking states that $\nabla f(0) = 0$ and I just don't get why.
I know that with $g(x) = \frac{||x||^2}{2}$ we have $\nabla g(0) = 0$ but how do I proceed with this error term $o(||x||^2)$.
I know that per def:Let $U \subseteq \mathbb{R}^{n}$, $g: U\to \mathbb{R}$, $x_{0} \in U$. Then $\sigma(g)$ contains the functions $f:U\to \mathbb{R}$, which satisfy:$$\lim_{ x \to x_{0}, x\neq x_{0} } \left|\frac{f(x)}{g(x)}\right|=0$$
Why is $\nabla f(0) = 0$?
EDIT: Because the comments are pain. I understand the approach provided in the comments with the expansion. I'm trying to understand the limes solution borught up in the answers. This is what I couold come up based on the answer why does this then imply $\nabla f(0) = 0$
$\lim_{x\to 0}\frac{f(x)-\nabla f \cdot(x)}{||x||^2} = \frac{o(||x||^2) -\nabla f \cdot(x)}{||x||^2} = \frac{o(||x||^2)}{||x||^2} - \frac{\nabla f \cdot(x)}{||x||^2} = -\frac{\nabla f \cdot(x)}{||x||^2}$