= Solution
Start from the <Erlang loss formula> and reverse the order in its denominator:
$$
\frac1{E(N\nu,NC)}=\sum_{j=0}^{NC}\frac{(NC)!}{(NC-j)!(N\nu)^j}=\sum_{j=0}^{NC}\prod_{\ell=0}^{j-1}\frac{NC-\ell}{N\nu}.
$$
For each fixed $j$, the product tends to $(C/\nu)^j$. If $\nu>C$, it is bounded by this same summable <geometric series>. Extend the summand by zero for $j>NC$ and apply the <dominated convergence theorem> with <counting measure> to obtain
$$
\frac1{E(N\nu,NC)}\longrightarrow\sum_{j=0}^\infty(C/\nu)^j=\frac1{1-C/\nu}.
$$
If $\nu\leq C$, retain the first $K+1$ summands. Their limiting sum is at least $K+1$, so the reciprocal tends to infinity as $K$ can be arbitrarily large. This includes $\nu=C$ without a separate normal approximation. Therefore
$$
\boxed{E(N\nu,NC)\longrightarrow\max\{0,1-C/\nu\}.}
$$
Here $C$ and $N$ are positive integers, as appropriate for circuit counts; taking $\lfloor NC\rfloor$ gives the same limit for a positive real capacity scale. This is the <proportional scaling limit of the Erlang loss formula>.
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