$\delta$ - method says that, suppose that
$$\frac{X_n-\mu}{\sigma_n}\xrightarrow{dist.}X$$
with $\mu\in\mathbb{R}$, $0<\sigma_n\rightarrow 0$ as $n\rightarrow \infty$ and $g:\mathbb{R}\rightarrow\mathbb{R}$ such that $g'(\mu)\neq 0$, then
$$ \frac{g(X_n) - g(\mu)}{\sigma_n} \xrightarrow{dist.} g'(\mu)X $$
now suppose I use the sequence of function ${g_n(x)}$ instead of $g(x)$, then what kind of modification needed to say something like usual $\delta$-method. Is there any result like that?
Thanks in advance
If you go through the proof of this result, you will see that basically it relies on the mean value theorem:
$g(X_n) - g(\mu) = g'(R_n)(X_n - \mu)$, where $R_n \in (X_n,\mu)\cup (\mu,X_n)$. In combination with $\sigma_n \to 0$, $X_n \stackrel{P}{\to} \mu$, the result comes out straightforwardly after dividing by $\sigma_n$.
If $g=g_n$, and $g_n$ is differentiable, you still have $g_n(X_n) - g_n(\mu) = g'_n(R_n)(X_n - \mu)$, so if $g_n$, $g_n'$ converge uniformly to some $g$ and $g'$, everything works out as before.