= Solution
The sum still has a <normal distribution>, now with
$$
\mathbb E\frac{C+S_L}{L}=\mu+\frac{\nu}{L},\qquad
\operatorname{Var}\frac{C+S_L}{L}=\frac{\sigma^2}{L}+\frac{\rho^2}{L^2}.
$$
Its scaled <cumulant-generating function> is
$$
\frac1L\log\mathbb E e^{\theta(C+S_L)}
=\mu\theta+\frac{\sigma^2\theta^2}{2}+\frac{\nu\theta+\rho^2\theta^2/2}{L}.
$$
The limit is the same everywhere-finite function as in part (b), so the <Gärtner–Ellis theorem> gives
$$
\boxed{I_C(x)=\frac{(x-\mu)^2}{2\sigma^2}}
$$
at <large-deviation speed> $L$, when $\sigma^2>0$.
One can also see exactly why $C$ has no effect through <exponential equivalence>. For every $\varepsilon>0$, the <Chernoff bound> for the two <Gaussian tail bounds> gives, when $L\varepsilon>|\nu|$ and $\rho^2>0$,
$$
\mathbb P(|C|>L\varepsilon)\leq2\exp\left(-\frac{(L\varepsilon-|\nu|)^2}{2\rho^2}\right).
$$
Its logarithm divided by $L$ tends to $-\infty$. If $\rho^2=0$, this probability is eventually zero. The transfer result for <exponential equivalence> says that coupled families whose distance exceeds each fixed positive tolerance with superexponentially small probability share a <large deviation principle> with the same <good rate function>. Apply it to $(C+S_L)/L$ and $S_L/L$. This also covers $\sigma^2=0$, with the point-mass <rate function> from part (b).
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