Convergence in distribution of a sequence of random variables, without using delta method Let $(A_n)_{n=1}^\infty$ be a sequence of random variable in $[0,1]$ and $A$ a constant in $(0,1)$ such that $\sqrt{n}(A_n - A)\overset{d}{\to}B$, where $B$ is a random variable and "$\overset{d}{\to}$" denotes convergence in distribution. Now, let $c$ be a constant such that $A \leq c < 1$. What can be said about the weak convergence of $\sqrt{n}(\min\{A_n, c\}- A)=\sqrt{n}(\min\{A_n, c\}- \min\{A, c\})\equiv \sqrt{n}(g(A_n)- g(A))$, with $g:[0,1]\mapsto [0,c]:x \mapsto\min\{x,c \}$? In the case of $A=c$, $g$ is not differentiable at $A$ and delta method can not be used to derive the weak limit of $\sqrt{n}(g(A_n)- g(A))$: how to proceed in that case?   
 A: By definition $\sqrt{n}(A_n - A)\overset{d}{\to}B$ iff 
$$
\mathbb P(\sqrt{n}(A_n - A) \leq x) \to F_B(x) 
$$
as $n\to\infty$ for any continuity point of $F_B(x)$. 
Consider arbitrary continuity point $x$ of $F_B(x)$. 
$$
\mathbb P(\sqrt{n}(\min(A_n, c) - A))\leq x) = \mathbb P\bigl(\min(A_n,c) \leq \frac{x}{\sqrt{n}}+A\bigr)$$ 
$$=1-\mathbb P\bigl(A_n > \frac{x}{\sqrt{n}}+A,\ c>\frac{x}{\sqrt{n}}+A\bigr) =1-\mathbb P\bigl(A_n > \frac{x}{\sqrt{n}}+A\bigr)\mathbb P\bigl(c>\frac{x}{\sqrt{n}}+A\bigr) 
$$ 
$$
=1-\bigl(1-\mathbb P(\sqrt{n}(A_n -A) \leq x)\bigr)\mathbb P(x < \sqrt{n}(c-A)) 
$$
If $c>A$ then $\sqrt{n}(c-A)\to+\infty$ and the last multiplier $P(x < \sqrt{n}(c-A))$ tends to $1$. For this case
$$
\mathbb P(\sqrt{n}(\min(A_n, c) - A))\leq x) \to F_B(x).
$$
We obtained for $c>A$ that 
$$
\sqrt{n}(\min(A_n, c) - A) \overset{d}{\to} B.
$$
If $c=A$ then $P(x < \sqrt{n}(c-A))=P(x < 0)$ equals to $1$ for $x<0$ and equals to $0$ for $x\geq 0$.
Then for $x\geq 0$
$$
\mathbb P(\sqrt{n}(\min(A_n, c) - A))\leq x) = 1,
$$
and for $x<0$
$$
\mathbb P(\sqrt{n}(\min(A_n, c) - A))\leq x) \to F_B(x).
$$
We obtained for $c=A$ that 
$$
\sqrt{n}(\min(A_n, c) - A) \overset{d}{\to} \min(B,0).
$$
