I don't know how to take the limit of the ratio of two functional error estimates, when both go to zero in particular $$ \lim_{h\to 0} \frac{h \| u - u_h\|^2_{L^2(\partial \kappa)} }{ \| u - u_h\|^2_{L^2(\kappa)} } $$
where $h$ is the diameter of $\kappa$ and $u_h = \arg \inf_{v\in \mathbb{P}^p} \|u - v\|_{L^2(\kappa)}$, and $\mathbb{P}^p(\kappa)$ is the set of polynomials of order less than or equal to $p$.
From computation I have pretty good reason to believe that this gives a constant value for sufficiently smooth $u$.
Right now I have, using the trace inequality, \begin{align} \lim_{h\to 0} \frac{h \| u - u_h\|^2_{L^2(\partial \kappa)} }{ \| u - u_h\|^2_{L^2(\kappa)} } &\leq \lim_{h\to 0} \frac{h \| u - u_h\|_{L^2(\kappa)}\| u - u_h\|_{H^1(\kappa)} }{ \| u - u_h\|^2_{L^2(\kappa)} }\\ &= \lim_{h\to 0} \frac{h \| u - u_h\|_{H^1(\kappa)} }{ \| u - u_h\|_{L^2(\kappa)} } \end{align}
but I am stuck after that, and am unsure if this is a sensible approach.
EDITS: $$\| u - u_h \|^2_{L^2(\kappa)} = \int_{\kappa} (u - u_h)^2 dx $$ $\kappa$ is a non-degenerate simplex, and $\partial\kappa$ is the boundary of $\kappa$.
Simplified 1D version: I am fairly numerically convinced the answer is $4(p+\frac{3}{2})$
$$\lim_{h\to 0} \frac{ h \left( (u-u_h)^2(x_M + \frac{h}{2}) +(u-u_h)^2(x_M - \frac{h}{2}) \right) }{ \int^{x_M + \frac{h}{2}}_{x_M - \frac{h}{2}} (u-u_h)^2 dx } $$
Note: This comes from looking for the value of $$ \frac{\partial}{\partial h} \log \left( \int^{x_M + \frac{h}{2}}_{x_M - \frac{h}{2}} (u-u_h)^2 dx \right) $$