I just want to make sure that my answer is correct.
I have $X_i \sim Poi(\lambda)$ and I need to find the asymptotic distribution of $\sqrt{n}(\overline{X}-\lambda)$ as $n\to\infty$
Central Limit Theorem:
$$\sqrt{n}\frac{\overline{X}_n - \mu}{\sigma} \xrightarrow[]{\;\;\;D\;\;\;} Z\sim(0,1) $$
$\mu = \lambda$ and $\sigma^2 = \lambda$ so
$$\sqrt{n}\frac{\overline{x}_n - \lambda}{\sigma} \approx Z \implies \sqrt{n}(\overline{x}_n - \lambda) \approx Z\sigma $$
and we know that $\mathbb{E}(Z) = 0$ so $\mathbb{E}(Z\sigma) = \sqrt\lambda \mathbb{E}(Z) = 0$ and $Var(Z\sigma) = \sigma^2Var(Z) = \lambda$ so in the end we have
$$\sqrt{n}(\overline{x}_n - \lambda) \sim N(0,\lambda)$$
Does this make sense?