$\text{Let}~(X_i)_{i=1}^n~ \text{be an i.i.d. sample of}$ $\text{n observations, with} ~ E(X_i)=\mu\in\mathbb{R}~ \text{and} ~ Var(X_i)=\sigma^2 \in (0, \infty)$
$\text{I'm asked to propose estimator}$ $\text{for the asymptotic variance of the}$ $\text{following distribution and to prove}$ $\text{that this estimator is consistent:}$ $\sqrt{n}\left(\ln \bar{X}_{n} - \ln \mu\right) \xrightarrow{d} \mathcal{N}(0,\frac{\sigma^2}{\mu^2})$
$\text{To do this, we have to use :}$ $\bar{X}_{n}=\frac{1}{n}\sum_{i=1}^{n}X_{i} ~\text{and}$
$B_n=\frac{1}{n}\sum_{i=1}^{n}X_{i}^{2}-\bar{X}_{n}^2$
$\text{I tried but didn't get something}$
Thanks!
$\bar{X}_n$ is an estimator of $\mu,$ $B_n$ is an estimator of $\sigma^2.$ A straightforward estimator for the asymptotic variance is then $$V_n = B_n/(\bar{X}_n^2).$$
Are you able to show that $B_n$ and $\bar{X}_n$ are consistent estimators of $\sigma^2$ and $\mu$ respectively?
If we have this consistency, we have by the continuous mapping theorem that also $\bar{X}_n^2$ is consistent for $\mu^2$ and we can then complete the proof that $V_n$ is consistent for $\sigma^2/\mu^2$ by using Slutsky's lemma.