= Solution
For a <functor> $D:\mathcal J\to\mathcal C$ with finite indexing <category>, form the finite <products in a category>
$$
P=\prod_{j\in\operatorname{Ob}\mathcal J}D(j),\qquad Q=\prod_{u:i\to j\text{ in }\mathcal J}D(j).
$$
Define $s,t:P\rightrightarrows Q$ by requiring the coordinate indexed by $u:i\to j$ to be $D(u)\pi_i$ for $s$ and $\pi_j$ for $t$. Let $e:L\to P$ be their <equalizer>, and put $\lambda_j=\pi_j e$. The equality $se=te$ says exactly that $D(u)\lambda_i=\lambda_j$ for every <morphism> $u$, so these <morphisms> form a <categorical cone>.
For any other <categorical cone> $(x_j:X\to D(j))$, the <product in a category> gives a unique $x:X\to P$ with $\pi_jx=x_j$. The <categorical cone> equations give $sx=tx$, so the <equalizer> gives a unique $\bar x:X\to L$ with $e\bar x=x$. This is exactly the <universal property> of a <categorical limit>. Empty indexing products are <terminal objects>, so the construction includes the empty <diagram in a category>. Thus \b[finite products and <equalizers> give all <finite limits>].
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