Let $f:\mathbb{R}^n \to \mathbb{R}$ continuous and differentiable in $\mathbb{R}^n \setminus \{0\}$ and for all $i \in \{ 1,...,n \}$ $$\lim_{h \to 0} \frac{\partial}{\partial x_i} f(h) = 0.$$ Show: Then $f$ is also differentiable in $0$.
I applied the definition of partial derivatives and got $$\lim_{h \to 0} \frac{\partial}{\partial x_i} f(h) = 0 \Longleftrightarrow \lim_{h \to 0}\left( \lim_{t \to 0} \frac{f(h + te_i) - f(h)}{t} \right) = 0$$ Taking the limit in the bracket yields $0$ and $h$ vanishes. Any help is appreciated.
for any $x \not= 0$ you have $$ \frac{f(x) - f(0)}{x} = f'(\alpha(x))$$ for some $\alpha(x)$ in the interval $(0,x)$. When $x$ goes to $0$ so does $\alpha(x)$ and so you're done, because the limit of the RHS exists and is equal to $0$ by your assumption. Hope you can fill in the tiny details yourself.
ok, let me elaborate some more: if I want to show that $f$ is differentiable at $0$ with derivative (gradient if you wish) equal to $0$ then I need to show precisely that $$ \lim_{|x| \rightarrow 0} \frac{f(x) - f(0)}{|x|} \rightarrow 0 \ldotp$$ What I do is I take arbitrary $x \not=0$ and I consider a function $g:[0,|x|] \rightarrow R$ (continuous and differentiable in the interior) defined by $$g(t) := f(\frac{t}{|x|} x)$$ then $$\frac{f(x) - f(0)}{|x|} = \frac{g(t) - g(0)}{t}$$ with $t = |x|$. standard mean value theorem for $g$ tells me then that there exists $\alpha \in [0,t] = [0, |x|]$ such that $$\frac{g(t) - g(0)}{t} = g'(\alpha)$$ but $$g'(\alpha) = [\nabla f](\frac{\alpha}{|x|} x) \cdot \frac{x}{|x|}$$ I need to show that $g'(\alpha) ( = g'(\alpha(x)))$ goes to $0$ as $x$ goes to $0$. For that I only need to show that $$ | [\nabla f](\frac{\alpha}{|x|} x) | \rightarrow 0$$ because I'm taking scalar product with a vector of norm $1$. Showing that the gradient of $f$ at this point goes to $0$ is easy, since the norm of the point in which I'm tkaing the gradient is equal to $|\alpha(x)|$ hence it's smaller than $|x|$. you know that all the partial derivatives tend to $0$, there are only finitely many of them hence they tend to $0$ uniformly (as a family) - I think you'll know how to show that part.