Asymptotic variance of an estimator of the parameter of esponential distribution. $**X**=\left(X_1,...X_n\right) $is a simple random sample from an exponential distribution $p\left(x;\theta\right)$, where $\theta$ is the parameter. $g_n\left(**X**\right)$ is an estimator of $\theta$. And $$\sqrt{n}\left(g_n\left(**X**\right)-\theta\right)\to N\left(0,V\left(\theta\right)\right).$$
So what is the relation between $V\left(\theta\right)$ and the Fisher information $I\left(\theta\right)$?

Well. By saying that $\sqrt{n}\left(g_n\left(X\right)-\theta\right)\xrightarrow d N\left(0,V\left(\theta\right)\right)$ you assume asymptotic normality of the estimator $g_n(\theta)$. If (in addition to normality) you say that $g_n(\theta)$ is a MVUE (or maybe just BLUE) (that means it is the best in some class of estimators), then hejseb is correct and you estimator will be asymptotically efficient and $V(g(\mathbf{X})) \to \frac{1}{\mathcal{I}[g(\mathbf{X})]}$. Otherwise there will be an estimator with the variance smaller than yours. But since your estimator is (at least) asymptotically unbiased then it will not be smaller than the one, given by the Cramer-Rao bound: $V(g(\mathbf{X}))\geq \frac{1}{\mathcal{I}[g(\mathbf{X})]}$.
