This is a homework problem. I have figured out part (a) but I need help with part (b). I include part (a) for completion.

Suppose $X_1,\ldots,X_n$ are iid Poisson random variables. Furthermore, let $Z_n$ be the proportion of zeroes observed i.e. $Z_n = n^{-1}\sum_{i=1}^n 1\{X_j=0\}$.

$(a)$ Find the joint asymptotic distribution of $\left(\bar{X}_n,Z_n\right)$

Since $\text{E}[X_1]=\theta$ and $\text{Var}[X_1]=\theta$, by the central limit theorem we have $$\sqrt{n}(\bar{X}_n-\theta) \overset{D}{\longrightarrow} Z_1,\quad Z_1\sim N(0,\theta)$$ and since $\text{E}[1\{X_1=0\}] = P(X_1=0) = e^{-\theta}$ and $$\text{Var}[1\{X_1=0\}] = \text{E}[1\{X_1=0\}^2] - \text{E}[1\{X_1=0\}]^2=e^{-\theta}-e^{-2\theta}=e^{-\theta}(1-e^{-\theta})$$ by the central limit theorem we have $$\sqrt{n}(Z_n-e^{-\theta}) \overset{D}{\longrightarrow} Z_2,\quad Z_2\sim N(0,e^{-\theta}(1-e^{-\theta}))$$ Furthermore we have $$\text{Cov}[X_1,1\{X_1=0\}] = 0 - \theta e^{-\theta}$$ Therefore, by the multivariate central limit theorem $$\sqrt{n}\begin{pmatrix} \bar{X}_n-\theta \\ Z_n - e^{-\theta}\end{pmatrix} \overset{D}{\longrightarrow} \mathbf{Y}, \quad \mathbf{Y} \sim \text{MVN}\left( \begin{pmatrix} 0 \\ 0 \end{pmatrix}, \begin{pmatrix} \theta & -\theta e^{-\theta}\\-\theta e^{-\theta} & e^{-\theta}\end{pmatrix}\right)$$

$(b)$ Based on your answer in (a), find the asymptotic distribution of $\sum_{i=1}^n X_i \big/ \sum_{i=1}^n 1\{X_i>0\}$. This is an estimate of the mean $\text{E}[X|X\geq 1]$ from a truncated Poisson.

We have $$\dfrac{\sum_{i=1}^n X_i}{\sum_{i=1}^n 1\{X_i>0\}}=\dfrac{n\bar{X}_n}{n-nZ_n} = \dfrac{\bar{X}_n}{1-Z_n}$$ I do not know how to proceed from here! I have a ratio of two normal distributions (marginal normal, and jointly normal).

$(c)$ Compute the exact mean and variance from a truncated Poisson$(\theta)$ with zero values truncated; i.e. $X\sim \text{Poisson}(\theta)$, compute $\text{E}[X|X\geq 1]$ and $\text{Var}[X|X\geq 1]$. Compare this to the asymptotic result in (b).