Let $g : \mathbb{R} \to \mathbb{R}$ be a differentiable function with $g(0) = 0$, $g'(0) = 0$ and $g'(1) \neq 0$. Let $f(x) = \begin{cases} \frac{x}{|x|}\, g(x), & x \neq 0 \\ 0, & x = 0 \end{cases}$ and $h(x) = e^{|x|}$ for all $x \in \mathbb{R}$. Let $(f \circ h)(x)$ denote $f(h(x))$ and $(h \circ f)(x)$ denote $h(f(x))$. Then which of the following is (are) true?

$g(0)=0,\quad g'(0)=0,\quad g'(1)\neq 0$

$f(x)=\dfrac{x}{|x|}\,g(x)\ (x\neq 0),\quad f(0)=0$

$h(x)=e^{|x|}$

$\text{Check differentiability at } x=0 \text{ of:}$

$f,\quad h,\quad f\circ h,\quad h\circ f$

$\text{Use } f'(0)=\lim_{x\to0}\dfrac{f(x)-f(0)}{x}$

$\text{Compare the left and right derivatives}$

$f'(0)=\lim_{x\to0}\dfrac{1}{x}\cdot\dfrac{x}{|x|}\,g(x)$

$x\to0^{+}:\ \dfrac{g(x)}{x}\to g'(0)=0,\qquad x\to0^{-}:\ -\dfrac{g(x)}{x}\to -g'(0)=0$

$\therefore f'(0)=0 \Rightarrow (1)\ \text{TRUE}$

$h'(0^{+})=1,\qquad h'(0^{-})=-1$

$1\neq -1 \Rightarrow h \text{ not differentiable at } 0 \Rightarrow (2)\ \text{FALSE}$

$(f\circ h)(x)=f(e^{|x|})=g(e^{|x|})$

$(f\circ h)'(0^{+})=g'(1),\qquad (f\circ h)'(0^{-})=-g'(1)$

$g'(1)\neq 0 \Rightarrow \text{not differentiable} \Rightarrow (3)\ \text{FALSE}$

$(h\circ f)(x)=e^{|f(x)|}=e^{|g(x)|}$

$g'(0)=0 \Rightarrow g(x)=o(x) \Rightarrow |g(x)|=o(|x|)$

$(h\circ f)'(0)=\lim_{x\to0}\dfrac{e^{|g(x)|}-1}{x}=0 \Rightarrow (4)\ \text{TRUE}$

$\boxed{\text{Answer: (1) and (4)}}$