Let H be an infinite dimensional Hilbert space and I_H be the identity operator on H. Then A = K(H)+C I_H isn't liminal (otherwise, since identity representation K(H)+C I_H® B(H) is nonzero irreducible ((K(H)+C I_H)¢ = K(H)¢ = CI_H (see CW17)) we should have I_H Î K(H), a contradiction). But K(H) is liminal, since each nonzero irreducible representation of K(H) is unitarily equivalent to identity representation K(H)® B(H) (see [Mur, page 146]). Also [ A/ J] @ CI_H which is finite dimensional and so is liminal (Every finite dimensional C^*-algebra B is liminal, since if (H₁,Y) is a nonzero irreducible representation of B, then for x ¹ 0 in H₁, Y(B)x is finite dimensional (for Y(B) is finite dimensional). If (u_l)_l be any approximate unit for B, then (Y(u_l))_l strongly converges to I_H so x Î [Y(B)x] = Y(B)x. Hence Y(B)x is a nonzero (closed) subspace of H₁ invariant for Y(B), so by irreducibility, Y(B)x = H₁. Therefore H₁ is finite dimensional. Thus Y(B) Í B(H₁) = K(H₁)).

Ref.
[Mur] G.J. Murphy, C^*-algebras and operator theory, Academic Press, 1990.