 Solution

Let $D:I\to \mathcal{C}$ be a finite diagram. Form the finite categorical productsThere are two morphisms $s,t:P\rightrightarrows Q$: their components at $u:i\to j$ are $D(u){\pi }_i$ and ${\pi }_j$, respectively. Let $e:L\to P$ be their equaliser. Then ${\ell }_i={\pi }_{i}e$ satisfy $D(u){\ell }_i={\ell }_j$, so they constitute a cone over a diagram.

Any other cone $x_i:X\to D(i)$ uniquely determines $x:X\to P$. Its cone equations say precisely that $sx=tx$. The equaliser therefore gives a unique $h:X\to L$ with $eh=x$, and thus ${\ell }_{i}h=x_i$. Conversely these component equations force $eh=x$ by the categorical product universal property, so there is exactly one such mediator. This proves that $(L,{\ell }_i)$ is a categorical limit of $D$.

The empty categorical products are included: if $I$ is empty, $P$ and $Q$ are terminal and the construction gives the empty categorical limit. Thus finite categorical products and equalisers give every finite categorical limit, as a special case of the construction of small limits from products and equalizers.