Step 1: Take the concrete case where $t_n = \bar{x}$, the sample mean, which is the efficient estimator of $\mu$ for i.i.d. observations with variance $\sigma^2$.
Step 2: Its variance is $\text{Var}(\bar{x}) = \sigma^2/n$. As $n \to \infty$, this ratio goes to 0, since $\sigma^2$ is a fixed finite constant while $n$ grows without bound.
Step 3: This rules out option (B), which wrongly claims the variance settles at $\sigma^2$, that would only be true for a single observation, not for $t_n$ with large $n$.
Step 4: Option (C) talks about convergence of $t_n$ to $\mu$, which is really a statement about consistency, not a direct consequence of the definition of efficiency being tested here, whereas the vanishing variance is exactly what efficiency plus the Cramer-Rao bound formula gives directly.
\[\boxed{\text{Var}(t_n) \to 0 \text{ as } n \to \infty}\]