= Solution
{c}
The possible limiting <correlation coefficients> are
$$
\boxed{-1\leq\rho\leq0.}
$$
The direction of the dependence comes from inversion of the same <counting process>: extra arrivals produce a larger count but an earlier $n$-th <Poisson arrival time>. Shared fluctuations therefore enter the two coordinates with opposite signs. Positive limiting <correlation> cannot arise; this can be formalized using <negative quadrant dependence of a Poisson count and an arrival time>.
At the same time scale, $f(n)=n$, part (iii) rules out every positive-positive quadrant. A <bivariate normal distribution> with standard <normal> <marginal distributions> and $\rho>-1$ assigns positive <probability> to such a quadrant. Thus, under the assumed joint limit, this choice forces $\rho=-1$. At very different time scales, $f(n)=n^4$, part (iv) gives $\rho=0$.
Intermediate time scales retain an intermediate fraction of the same arrival fluctuations. In particular, the <Poisson count-arrival joint central limit theorem> gives $\rho=-1/\sqrt\lambda$ for $f(n)=\lambda n$ with $1\leq\lambda<\infty$. These choices realize every $\rho\in[-1,0)$, and $f(n)=n^4$ supplies zero. This identifies the whole interval without requiring an additional limit calculation in this part.
The premise about a <bivariate normal distribution> is understood to apply when a joint limit exists. Merely requiring $f(n)\to\infty$ cannot guarantee one: taking $f(n)=n$ for even $n$ and $f(n)=n^4$ for odd $n$ gives different limits along the two <subsequences> just identified.
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