To complete Michael's answer, the only situation that is under control is that of the Cauchy problem: the spatial domain is ${\mathbb R}^d$ or ${\mathbb T}^d$ (case of periodic solutions). This means that there is no boundary condition.
If $d=2$, both systems are globally well-posed for $t>0$, with uniformly bounded (in $L^2$) solutions, and $u_\nu$ converges strongly to the solution of the Euler equation. Notice that it is not a trivial fact: the reasons why both Navier-Stokes and Euler Cauchy problems are globally well-posed have nothing in common; for Navier-Stokes, it comes from the Ladyzhenskaia inequality (say, $H^1\subset L^4$), while for Euler, it is the transport of the vorticity.
If $d=3$, both Cauchy problems are locally-in-time well-posed for smooth enough initial data. One has a convergence as $\nu\rightarrow0+$ on some time interval $(0,\tau)$, but $\tau$ might be strictly smaller than both the time of existence of Euler and the $\lim\inf$ of the times of existence for Navier-Stokes.
To my knowledge, the initial-boundary value problem is a nightmare. The only result of convergence is in the case of analytic data (Caflisch & Sammartino, 1998). From time to time, a paper or a preprint appears with a `proof' of convergence, but so far, such papers have all be wrong.
By the way, your question is incorrectly stated, when you say boundary conditions are equal. The boundary condition for NS is $u=0$, whereas that for Euler is $u\cdot\vec n=0$, where $\vec n$ is the normal to the boundary. This discrepancy is the cause of the boundary layer. One may say that the difficulty lies in teh fact that this boundary layer is characteristic. Non-characteristic singular limits are easier to handle.
Another remark is that some other boundary condition for NS are better understood. For instance, there is a convergenece result (Bardos) when $u=0$ is replaced by $$u\cdot\vec n=0,\qquad {\rm curl}u\cdot\vec n=0.$$