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$f:\mathbb{R}^{2}\to\mathbb{R}$ , we're given that for any $ v\in\mathbb{R}^{2}$ the directional derivative $ \nabla_{v}f\left(0,0\right)$ exists.

1) Is $f$ continuous at $ \left(0,0\right)?$

2) Additionally, if we assume that $f$ is continuous at $ \mathbb{R}^{2}\setminus\left\{ 0,0\right\}$ , is $f$ continuous at $ \left(0,0\right)?$

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See here for an example showing that both questions have negative answers. – David Mitra Dec 4 '12 at 14:00
up vote 2 down vote accepted

Draw around each point $(x_n,y_n):=\left({1\over n},{1\over n^2}\right)$ $\ (n\geq 1)$ a small circle of radius ${1\over n^3}$ and erect on this circle a cone of hight $1$ having its tip at $(x_n,y_n,1)$. Let the graph of $f$ consist of the union of these conical surfaces, and add to that the plane minus the small circles. Then $f$ is continuous on ${\mathbb R}^2\setminus\{(0,0)\}$, but not at $(0,0)$. Furthermore all directional derivatives $\nabla_v f(0,0)$ exist and are $=0$.

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