$f \colon \mathbb R^n \to \mathbb R$ is continuous everywhere, and continuously differentiable on $\mathbb R^{n} \setminus \{0\}$.
We are given $\lim _{x \to 0} \nabla f(x) = L \in \mathbb R^n$.
Show that $f$ is $C^1$.
This is equivalent to showing that $\nabla f(0) = L$.
I've managed to do this using single variable calculus MVT. Basically show that each coordinate of the gradient individually is equal to $L_i$.
Is there a way to do this using the multivariable MVT?
Let $\delta \in \mathbb R^n$.
$f(\delta) - f(0) = \langle \nabla f(\xi), \delta \rangle$ where $\xi$ is in the straight line connecting vectors $0$ and $\delta$.
I can't divide by $\delta$ and isolate $\nabla f(\xi)$ as it is a vector. Unlike the single variable case. Is there a way to make this argument work or do I actually have to go through the single variable cases?
My solution, in case you are interested (not relevant to the question):
Define $g_i\colon \mathbb R \to \mathbb R$ by $g_i(x)= f(0,0, \dots ,0, x,0,\dots, 0)$, zero everywhere except the $i$'th coordinate.
Then $\nabla f(0)_i =\lim _{\delta \to 0}\frac{g_i(\delta)-g_i(0)}{\delta} = \lim _{\delta \to 0} g_i'(\xi)$ where $\xi \in (0, \delta)$
As $\delta$ approaches zero, $\xi$ must approach zero as well, so this is equal to $\lim_{\xi \to 0} g_i'(\xi)$ which we were given is equal to $L_i$, and we are done.