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Let f:\mathbb{R}^n \rightarrow \mathbb{R} be differentiable on \mathbb{R}^n \setminus \{0 \} and continuous at 0 . If

    \[\lim_{x \rightarrow 0} \frac{\partial f}{\partial x_i} (x) =0\]

for i=1, 2, \dots , n then prove that f is differentiable at 0.


Let us assume without loss of generality that f(0)=0. We will show that f(x) = o \left ( \left \| x \right \|_{\infty} \right ) as x \rightarrow 0 and that means that f is differentiable at x=0 with f'(0)=0.

Fix \epsilon>0 and choose \delta \in \left( 0 , \frac{1}{n}\right) such that 0< \left \| x \right \|_{\infty} < \delta1 \leq i \leq n. Therefore \left | f_{x_i} \right |<\epsilon. Now suppose \left \| x \right \|_{\infty}<\delta. Let a_k\; ,  \; 0 \leq k \leq n be the point that coincides with x on the first k coordinates and is 0 elsewhere. Then a_0,\dots,a_n is a path from 0 to x and each vector a_{k+1}-a_k is parallel to one of the axes. Hence

    \begin{align*} \left | f(x) \right |&=\left| \sum_{k=0}^{n-1}(f(a_k)-f(a_{k-1}) \right|\\ &\leq \sum_{k=0}^{n-1} \left|f(a_k)-f(a_{k-1}) \right |\\ &\leq n\epsilon \left \|a_k-a_{k-1}| \right \|\\ &<n\delta \epsilon \\ &<\epsilon \end{align*}

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