Let $f(x,y) = \begin{cases} 1, & \quad if \: xy=0, \\ xy, & \quad xy \neq 0. \end{cases}$ Then $f$ is continuous at $(0,0)$ and $\frac{\partial f}{ \partial x} (0,0)$ exists $f$ is not continuous at $(0,0)$ and $\frac{\partial f}{ \partial x} (0,0)$ exists ... $f$ is not continuous at $(0,0)$ and $\frac{\partial f}{ \partial x} (0,0)$ does not exist

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Jun 2, 2016
in Calculus
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