= Solution
Choose a limit $L_j$ for each diagram $F(-,j)$, with legs $p_{i,j}:L_j\to F(i,j)$. For an arrow $v:j\to j'$, the maps $F(1_i,v)p_{i,j}$ form a cone over $F(-,j')$, since the two-variable <functor> respects the commuting arrows in $I\times J$. The limit property therefore gives a unique arrow $L(v):L_j\to L_{j'}$ satisfying
$$
p_{i,j'}L(v)=F(1_i,v)p_{i,j}\quad\text{for every }i.
$$
The arrows $L(1_j)$ and $1_{L_j}$ have identical composites with all projections, so they are equal. The same uniqueness shows $L(v'v)=L(v')L(v)$. Hence the assignment extends to a <functor> $L:J\to\mathcal D$.
For the chosen limit objects and cones it is uniquely determined, literally, by these equations. Changing the choices gives a unique family of <isomorphisms> between the limits compatible with their projections; the equations above show that this family is a <natural isomorphism> of the resulting <functors>. Thus it is \b[unique up to a unique cone-compatible natural <isomorphism>], not independent of choices as an equality of objects.
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