Let $f : \mathbb{R} \to \mathbb{R}$ and $g : \mathbb{R} \to \mathbb{R}$ be functions satisfying $f(x+y) = f(x) + f(y) + f(x)f(y)$ and $f(x) = xg(x)$ for all $x, y \in \mathbb{R}$. If $\lim_{x \to 0} g(x) = 1$, then which of the following statements is/are True?

$f(x+y) = f(x) + f(y) + f(x)f(y)$ $f(x) = xg(x),\qquad \lim_{x\to 0} g(x) = 1$ $\text{Check: } f \text{ differentiable? } g \text{ diff. if } g(0)=1?\ f'(1)=1?\ f'(0)=1?$ $1.\ h(x)=1+f(x) \Rightarrow \text{multiplicative}$ $2.\ f(0),\ h(0)$ $3.\ \lim \Rightarrow h(x) \Rightarrow \text{check options}$ $h(x) = 1 + f(x) \Rightarrow h(x+y) = h(x)h(y)$ $f(0) = 0\cdot g(0) = 0 \Rightarrow h(0) = 1$ $f'(0) = \lim_{x\to0}\frac{f(x)-f(0)}{x} = \lim_{x\to0}\frac{f(x)}{x} = \lim_{x\to0}g(x)$ $= 1 \Rightarrow (4)\ \text{TRUE}$ $\lim_{x\to0}h(x) = 1 \Rightarrow h \text{ continuous, multiplicative}$ $h(x) = e^{h'(0)x} = e^{x}\quad (h'(0)=f'(0)=1)$ $f(x) = e^x - 1 \Rightarrow f \text{ differentiable everywhere} \Rightarrow (1)\ \text{TRUE}$ $f'(x) = e^x$ $f'(1) = e \neq 1 \Rightarrow (3)\ \text{FALSE}$ $x\neq0:\ g(x) = \frac{e^x-1}{x}$ $g(0)=1 \Rightarrow g \text{ differentiable everywhere} \Rightarrow (2)\ \text{TRUE}$ $\text{True: (1), (2), (4)}$ $\boxed{\text{Answer: (1), (2), (4)}}$