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The Laplacian operator in cylindrical coordinates is given by $$\Delta = \frac{\partial^2}{\partial r^2} +\frac{1}{r} \frac{\partial}{\partial r} + \frac{1}{r^2}\frac{\partial^2}{\partial \theta^2} + \frac{\partial^2}{\partial z^2}$$

Is there a limit that the Laplacian operator tends to as $r \to 0$?

Similarly, the divergence of a vector(say velocity vector) is given in cylindrical coordinates by $$\nabla \cdot \textbf{u} = \frac{\partial u_r}{\partial r} + \frac{u_r}{r} + \frac{1}{r}\frac{\partial u_{\theta}}{\partial \theta} + \frac{\partial u_z}{\partial z}$$

What is the limits of this divergence as $r\to 0$?

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"The limit of Laplacian as $r\to 0$" (whatever that means) is $$ \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} $$ Actually, it's not just at $r=0$ but everywhere. And for divergence it is $$ \frac{\partial u_x}{\partial x} + \frac{\partial u_y}{\partial y} + \frac{\partial u_z}{\partial z} $$

The point is, there is nothing special about Laplacian or divergence at $r=0$. There is something special about cylindrical coordinates there: $\theta$ is underfined, so one can't use cylindical coordinates to express differential operators when $r=0$.

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