Let $\mathbb{R}$ denote the set of all real numbers. Let $f : \mathbb{R} \to \mathbb{R}$ be an arbitrary function and let $g : \mathbb{R} \to \mathbb{R}$ be the function defined by $g(x) = x f(x)$, for all $x \in \mathbb{R}$. Then which of the following statements is (are) TRUE?

$g(x) = x\,f(x) \quad \forall\, x \in \mathbb{R}, \qquad g(0) = 0 \cdot f(0) = 0$ $\text{Which of (1), (2), (3), (4) are TRUE?}$ $1.\ \dfrac{g(x)-g(0)}{x-0} = f(x) \quad (x \neq 0)$ $2.\ \text{test (1)-(4); counterexamples for the false ones}$ $\text{For } x \neq 0: \quad \dfrac{g(x)-g(0)}{x-0} = \dfrac{x\,f(x)}{x} = f(x)$ $(1)\ \text{FALSE. Take } f(x)=\dfrac{1}{x}\ (x\neq0), \ f(0)=0$ $\Rightarrow g(x)=1 \ (x\neq0), \ g(0)=0 \ \Rightarrow \text{ jump at } 0$ $(2)\ \text{TRUE. } f \text{ continuous at } 0 \Rightarrow f(x) \to f(0)$ $\Rightarrow g'(0) = \lim_{x \to 0} \dfrac{g(x)-g(0)}{x} = f(0) \text{ exists}$ $(3)\ \text{FALSE. Take } f(0)=1, \ f(x)=0 \ (x\neq0)$ $\Rightarrow g \equiv 0 \text{ (differentiable)}, \text{ but } f \text{ not continuous at } 0$ $(4)\ \text{TRUE. } g \text{ differentiable at } 0 \Rightarrow f(x) = \dfrac{g(x)}{x} \to g'(0)$ $\Rightarrow \lim_{x \to 0} f(x) = g'(0) \text{ exists}$ $\boxed{\text{TRUE: statements } (2) \text{ and } (4)}$