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
$$u\cdot\vec n=0,\qquad {\rm curl}u\cdot\vec n=0.$$