Suppose $g:R\rightarrow R$ has at least three bounded continuous derivatives and let $X_i$ be $iid$ and in $L^2$. Prove that:
$\sqrt{n}[g(\overline{X_n}) - g(\mu)]\xrightarrow{w} N(0,g^{'}(\mu)^{2} \sigma ^2)$ i.e. weak convergence happens
where $\overline{X_n} = \frac{\sum X_n}{n}$, $\mu = EX_1$, $\sigma^2=Var(X_1)$
My attempt:
Let $\phi$ be the characteristic function/fourier transform. Weak convergence $\iff$ convergence of $\phi$
On the RHS, $N(0,g^{'}(\mu)^{2} \sigma ^2) \sim g^{'}(\mu) \sigma N(0,1) = g^{'}(\mu) \sigma \mathscr{X}$ where $\mathscr{X}$ is standard normal
$\phi_{g^{'}(\mu) \sigma \mathscr{X}}(t) = \phi_{\mathscr{X}}(g^{'}(\mu) \sigma t) = e^{-\frac{g^{'}(\mu)^{2} \sigma ^2 t^2}{2}}$ because $\phi_{\mathscr{X}}(t) = e^{-\frac{t^2}{2}}$
On the LHS:
The characteristic function is
$E[e^{it\sqrt{n}g(\overline{X_n}) - it\sqrt{n}g(\mu)}] = e^{-it\sqrt{n}g(\mu)} E[e^{it\sqrt{n}g(\overline{X_n})}]$
Because RHS has $g^{'}(\mu)$, I am thinking of expanding $e^{it\sqrt{n}g(\overline{X_n})}$ using Taylor's which is:
$e^{it\sqrt{n}g(\overline{X_n})} \approx e^{it\sqrt{n}g(\mu)} + e^{it\sqrt{n}g(\mu)}it\sqrt{n}g'(\mu)(\overline{X_n}-\mu)+e^{it\sqrt{n}g(\mu)}it\sqrt{n}g'(\mu)[(it\sqrt{n}g'(\mu))^2 + it\sqrt{n}g^{''}(\mu)](\overline{X_n}-\mu)^2$
So $LHS = e^{-it\sqrt{n}g(\mu)} E[e^{it\sqrt{n}g(\overline{X_n})}] = 1+[(it\sqrt{n}g'(\mu))^2 + it\sqrt{n}g^{''}(\mu)]E[\overline{X_n}-\mu]^{2} =$
$1+[(it\sqrt{n}g'(\mu))^2 + it\sqrt{n}g^{''}(\mu)](\frac{\sigma^2}{n}+\mu^2)$
I am stuck at this point as $RHS$ has $e$ to the some power which means $LHS$ needs to have a $(1+x/n)^n$ kind of a situation which I don't. Basically, I am missing a power $n$ on the $LHS$ which should occur because of $\overline{X_n}$ but $g(\overline{X_n})$ kind of blocks that to happen. A hint is appreciated about the whole approach, thanks.
By the strong law of large numbers, we have $$\lim_{n\to\infty}\frac{g(\overline{X_n}) - g(\mu)}{\overline{X_n}-\mu} = g'(\mu)$$ almost surely. By the central limit theorem and Slutsky's theorem, it follows that $$\sqrt{n}[g(\overline{X_n}) - g(\mu)] = \frac{g(\overline{X_n}) - g(\mu)}{\overline{X_n}-\mu}\cdot \sqrt{n} (\overline{X_n} - \mu) \xrightarrow{w} g'(\mu)X, $$ where $X \sim N(0,\sigma^2).$