= Solution
Let $D:J\to\mathcal C$ be a finite <diagram in a category>. Form the finite <products in a category>
$$
P=\prod_{j\in\operatorname{Ob}J}D(j),\qquad Q=\prod_{a:j\to k\ \text{in }J}D(k).
$$
Define $s,t:P\rightrightarrows Q$ by $\pi_a s=D(a)\pi_j$ and $\pi_a t=\pi_k$. Take their <equalizer> $e:L\to P$. The maps $\lambda_j=\pi_j e$ satisfy $D(a)\lambda_j=\lambda_k$, so they constitute a <categorical cone>.
Any other <categorical cone> $x_j:X\to D(j)$ gives a unique $x:X\to P$ by the <product in a category> property. Its cone equations say $sx=tx$, so the <equalizer> gives a unique $\bar x:X\to L$ with $e\bar x=x$. The projections then give $\lambda_j\bar x=x_j$, and uniqueness follows from both universal properties. Thus \b[this <equalizer> of two product maps is the <finite limit> of the <diagram in a category>].
The empty indexing <category> is included: its empty <product in a category> is a <terminal object>, which is the empty <categorical limit>. Hence no nonemptiness restriction has entered the construction.
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