I am trying to solve the following:
Let $X_1, X_2, . . .$ be i.i.d. r.v.s with mean $\mu$ and positive, finite variance $\sigma^2$, and set $Sn = \sum_{k=1}^{n} X_k, n ≥ 1$. Suppose that $g$ is twice continuously differentiable, and that $g'(\mu)$ is non zero. Show that $\sqrt{n}(g(\frac{S_n}{n})-g(\mu)) \to N(0,b^2)$ in distribution and determine $b^2$.
So far I started by taking the taylor series of $g(x)$ around $\mu$. So $g(x)=g(\mu)+g'(\mu)(x-\mu)+...$ Letting $x=\bar{x}=\frac{S_n}{n}$ the sample average, we have $g(\bar{x})=g(\mu)+g'(\mu)(\bar{x}-\mu)+...$
By rearranging terms we have $g(\bar{x})-g(\mu)=g'(\mu)(\bar{x}-\mu)$. Substituting this into the initial problem, we have $\sqrt{n}(g'(\mu)(\bar{x}-\mu))=\sqrt{n}(g'(\mu)(\frac{S_n}{n}-\mu))$.
At this point it is almost in a form where I can apply the CLT, but I'm not sure how to rearrange with the $g'(\mu)$ factor in order to get into a form to apply the theorem and determine $b^2$.
Let $\bar X_n=n^{-1}S_n$. By the CLT you have
$$\sqrt{n}(\bar X_n-\mu)\rightarrow N(0,\sigma^2)$$
Now, apply the mean value theorem to get
$$g(\bar X_n)-g(\mu)=g'(\bar X_n')(\bar X_n-\mu)$$
where $\bar X_n'$ lies between $\bar X_n$ and $\mu$ so that $g'(\bar X_n')\rightarrow^p g'(\mu)$ (by the WLLN and the continuous mapping theorem) and
$$\sqrt{n}(g(\bar X_n)-g(\mu))= \left[g'(\mu)+o_p(1)\right]\sqrt{n}(\bar X_n-\mu)\rightarrow^dN(0,[g'(\mu)]^2\sigma^2)$$