= Border rank
{title2=$\underline R(T)$}
Over $\mathbb C$, the <border rank> of a <tensor> $T$ is the least $r$ for which $T$ lies in the closure of tensors of <tensor rank> at most $r$. The usual <Euclidean topology> and <Zariski topology> give the same closure here. Hence $\underline R(T)\le R(T)$, but equality need not hold.<Geometry and complexity theory by J. M. Landsberg>[tensor rank, border rank and secant varieties]{ref}
Let $e_0,e_1$ be a basis of $\mathbb C^2$ and put
$$
T=e_1\otimes e_0\otimes e_0+e_0\otimes e_1\otimes e_0+e_0\otimes e_0\otimes e_1.
$$
For nonzero $\varepsilon$,
$$
\frac{(e_0+\varepsilon e_1)^{\otimes3}-e_0^{\otimes3}}{\varepsilon}\longrightarrow T.
$$
Each approximant is a sum of two <pure tensors>, so the border rank is at most two. It is not one, since a matrix obtained by separating one factor from the other two has rank two.
Its exact <tensor rank> is three. The two <tensor slices> in the first factor span $\left\{\left(\begin{smallmatrix}b&a\\a&0\end{smallmatrix}\right):a,b\in\mathbb C\right\}$. Its determinant is $-a^2$, so this space has only one rank-one direction. A rank-two tensor with two independent slices would require their span to be generated by two independent rank-one matrices, a contradiction. The limit therefore permits a decomposition that is impossible exactly.
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