For the solution of 
$$
\begin{cases}
\lambda u^\epsilon - \frac{\epsilon^2}{2} \Delta u^\epsilon = 0 &\text{in } \Omega \\
u^\epsilon=1 & \text{on } \partial \Omega
\end{cases}
$$
Varadhan proved that 
$$\lim_{\epsilon \to 0} - \epsilon \log u^\epsilon = \sqrt{2\lambda} \mathrm{dist} (x,\partial \Omega) $$


Can we prove that something similar for 
$$
\begin{cases}
\lambda u^\epsilon + \frac{\epsilon^2}{2} (-\Delta)^s u^\epsilon = 0 &\text{in } \Omega \\
u^\epsilon=1 & \text{on } \mathbb R^N \setminus \Omega
\end{cases}
$$
holds?
Here $(-\Delta)^s$ is the [fractional Laplacian][1].


  [1]: https://en.wikipedia.org/wiki/Fractional_Laplacian