On the empirical mean and variance of a Poisson i.i.d. sample

Let $X_1, X_2, \ldots, X_n$ be a random sample from a Poisson($\lambda$) distribution. Let ($\bar{X}$) be their sample mean and $s^2$ their sample variance.

1. Show that $\frac{\sqrt{n}[\bar{X}-\lambda]}{\sqrt{\bar{X}}}$ and $\frac{\sqrt{n}[\bar{X}-\lambda]}{s}$ both have a standard normal limiting distribution.
2. Find the limiting distribution of $\sqrt{n}[\bar{X}-\lambda]^2$
3. Find the limiting distribution of $\sqrt{n}[\bar{X}^2-\lambda^2]$

1) For $\frac{\sqrt{n}[\bar{X}-\lambda]}{S}$, we know that $\frac{\sqrt{n}[\bar{X}-\lambda]}{S}$ = $(\frac{\sqrt{n}[\bar{X}-\lambda]}{\sigma})(\frac{\sigma}{S})$. Since $\frac{\sqrt{n}[\bar{X}-\lambda]}{\sigma}$ appraches N(0,1) in distribution by CLT, and since $(\frac{\sigma}{S})$ appraches 1 in probability, the whole thing approaches N(0,1). For $\frac{\sqrt{n}[\bar{X}-\lambda]}{\sqrt{\bar{X}}}$, we get the same result...since the mean is equal to the variance in a poisson distribution.

3) From a theorem in my textbook, I know that if $\sqrt{n}(X_n - \theta) \rightarrow N(0,\sigma^2)$ and if there is a differentiable function g(x) at theta where the derivative at theta is not zero...then $\sqrt{n}(g(X_n)-g(\theta)) \rightarrow N(0,\sigma^2(g'(\theta))^2)$. So I just need to use this theorem, right? And in this case $g(x)=x^2$.