Let $\mathbb{R}$ denote the set of all real numbers. Define the function $f$: $\mathbb{R} \to \mathbb{R}$ by $f(x) = \begin{cases} 2 - 2x^{2} - x^{2}\sin\frac{1}{x} & \text{if } x \neq 0, \\ 2 & \text{if } x = 0 \end{cases}$. Then which one of the following statements is TRUE?

$f(x) = 2 - 2x^2 - x^2\sin\tfrac1x\ (x\neq 0),\qquad f(0) = 2$

$2 - f(x) = x^2\left(2 + \sin\tfrac1x\right)$

$\text{To find: the TRUE statement among } (1)\text{–}(4)$

$\lim_{x\to0}\dfrac{f(x)-f(0)}{x} = \lim_{x\to0}\left(-2x - x\sin\tfrac1x\right) = 0$

$f'(0) = 0 \Rightarrow f \text{ IS differentiable at } 0 \Rightarrow (1)\ \text{FALSE}$

$f'(x) = -4x - 2x\sin\tfrac1x + \cos\tfrac1x \quad (x\neq 0)$

$x\to0:\ -4x-2x\sin\tfrac1x \to 0,\quad \cos\tfrac1x \text{ oscillates}$

$f' \text{ changes sign arbitrarily near } 0$

$\Rightarrow f \text{ is monotone on NO interval around } 0$

$(2)\ \exists\,\delta>0:\ f \text{ decreasing on } (0,\delta)?\ \text{NO} \Rightarrow \text{FALSE}$

$(3)\ \forall\,\delta>0,\ f \text{ not increasing on } (-\delta,0) \Rightarrow \text{TRUE}$

$2 - f(x) = x^2\left(2+\sin\tfrac1x\right) > 0 \text{ for } x\neq0$

$\Rightarrow f(x) < f(0):\ 0 \text{ is a local maximum} \Rightarrow (4)\ \text{FALSE}$

$\boxed{\text{option } (3)}$