I'd like to show that \begin{equation} \sqrt{n} \left( (1-\frac{1}{n})^{n\bar{X}} - e^{-\bar{X}} \right) \to 0 \end{equation} in probability for a random variable with mean $\mu$ and finite variance $\sigma^2$.
I believe this could be done with Taylors theorem though I'm not sure. Notice we can show using the continuous mapping theorem plus the law of large numbers that the quantity inside the paraenthesis converges to 0 in probability. The thing I can't get is how to show that even if we multiply it by $\sqrt{n}$ it still goes to 0.
There is probably a sign error in the problem, hence we consider $$Y_n=\left(1-\frac1n\right)^{n\bar X_n}-\mathrm e^{-\bar X_n}.$$ Note that, when $n\to\infty$, $$\log\left(1-\frac1n\right)=-\frac1n-\frac1{2n^2}+o\left(\frac1{n^2}\right),$$ hence $$\left(1-\frac1n\right)^{n\bar X_n}=\exp\left(-\bar X_n-\frac{\bar X_n}{2n}+o\left(\frac{\bar X_n}n\right)\right),$$ Thus, $$n\,Y_n\,\mathrm e^{\bar X_n}=n\,\left(\exp\left(-\frac{\bar X_n}{2n}+o\left(\frac{\bar X_n}n\right)\right)-1\right)=-\frac12\bar X_n+o\left(\bar X_n\right).$$ Note that $(\bar X_n)$ is almost surely bounded and that, by the strong law of large numbers for i.i.d. sequences, $\bar X_n\to\mu$ almost surely. Thus, almost surely, $$n\,Y_n\to-\frac12\mu\,\mathrm e^{-\mu}.$$ In particular, for every $\alpha\lt1$, $n^\alpha\,Y_n\to0$ almost surely, hence in probability.