Normal derivative Let $f(x,y)=\frac{1}{2}\log(x^2+y^2)$ in $\mathbb{R}^2\setminus (0,0)$. Is the limit of normal derivative of $f$ with respect to the unit circle ($\mathbb{S}=\{(x,y): \: x^2+y^2=1\}$) as $(x,y)$ approaches to a point on $\mathbb{S}$ equal to $1$?
 A: Changing to polar coordinates, the gradient of f is 
$$\nabla f=\frac{\partial f}{\partial r}\mathbf{e}_r+\frac{1}{r}\frac{\partial f}{\partial \theta}\mathbf{e}_\theta = \frac{\partial }{\partial r}\big{[}\frac{1}{2}\log(r^2)\big{]} \mathbf{e}_r= \frac{1}{r} \mathbf{e}_r.$$
Consequently, at any point on $S$ where $r=1$ and the outward unit normal vector is $\mathbf{n}=\mathbf{e}_r$, the normal derivative of $f$ is
$$\frac{\partial f}{\partial n} \Big{|}_S = \nabla f \cdot \mathbf{n}= \mathbf{e}_r \cdot \mathbf{e}_r=1$$
The gradient is continuous on $\mathbf{R}^2 \setminus (0,0)$ and approaches $\mathbf{e}_r$ at as we approach a point on $S$.
However, a normal derivative depends not only on the gradient but on a reference curve.  A question about the limit of the normal derivative as a point $\mathbf{x} \notin S$ approaches a point $\mathbf{x}_S \in S$ is somewhat vague. The answer depends on how the normal derivative is defined at the general point $\mathbf{x}$ and exactly how the limiting process is specified. 
At most, we can say that the normal derivative of $f$ at $\mathbf{x}=(r,\theta)$ is 
$$\frac{\partial f}{\partial n}(\mathbf{x})=\frac{\alpha}{r}$$
where $|\alpha|\leq1$ and the particular value of $\alpha$ depends on the orientation of a specific reference curve $C$ containing $\mathbf{x}$.
Then
$$\lim_{\mathbf{x} \rightarrow \mathbf{x}_S} \frac{\partial f}{\partial n}(\mathbf{x})=\alpha.$$
If we constrain the limiting process so that the normal vector of the reference curve approaches $\mathbf{e}_r$ then we have the limit equal to $1$. 
