Fractional Laplacian 
Q(1) It is well known that if $s \rightarrow 1$ then the fractional Laplacian converges to the classical Laplacian but the form of the Laplacian still remains in the non-local form whereas it is known that the Laplacian is a local differential operator. 

Why is this?.
(2) Also why do we choose the boundary condition as $u=0$ $\in \mathbb{R}^N\setminus\Omega$ and not just in $\partial\Omega$, if the domain $\Omega$ is bounded in $\mathbb{R}^N$.
 A: The expression you are referring to (I believe) is
$$\Delta^su(x) =c_{n,s} \int_{\mathbb{R}^n} \frac{u(y) - u(x)}{|x-y|^{n+2s}} \, dy,$$
which holds for $0 < s < 1$. Here, $c_{n,s}$ is a constant depending on $n$ and $s$. The expression does not hold for $s=1$; in fact it is not even integrable. Consider for example $u(x) = |x|^2$ so we have
$$\Delta^s u(0) = c_{n,s}\int_{\mathbb{R}^n} \frac{1}{|y|^{n + 2(s-1)}} \, dy.$$
When $s=1$, the integrand is $1/|y|^n$, which not integrable, and the principal value of the integral is infinity.
However, as you said, as $s\to 1^-$, $\Delta^su(x) \to \Delta u(x)$. What is happening is that $c_{n,s} \to 0^+$ as $s\to 1^-$, the integral is blowing up as $s\to 1^-$, and the two cancel each other out perfectly to give you $\Delta u(x)$. 
One way you can view this is that when $s<1$ is very close to $1$, you are averaging with the kernel $c_{n,s}/|y|^{n+2(s-1)}$, which is very close to being not integrable at $y=0$. Hence, it puts an enormous amount of its weight near $y=0$, and the closer $s$ is to $1$, the more of its weight (proportionally) is concentrated at $y=0$. In the limit as $s\to 1^-$ it becomes local, though the integral expression is no longer valid. You could make the expression valid by writing
$$\Delta u(x) = \lim_{s\to 1^-} c_{n,s} \int_{\mathbb{R}^n} \frac{u(y) - u(x)}{|x-y|^{n+2s}}\, dy.$$
A shorter explanation: If you average with a kernel that has infinite mass, regions with finite mass are negligible. 
