Let $f: \mathbb{R} \to (0, \infty)$ and $g: \mathbb{R} \to \mathbb{R}$ be twice differentiable functions such that $f''$ and $g''$ are continuous functions on $\mathbb{R}$. Suppose $f'(2) = g(2) = 0$, $f''(2) \neq 0$ and $g'(2) \neq 0$. If $\lim_{x \to 2} \frac{f(x)g(x)}{f'(x)g'(x)} = 1$, then

Let f: R → (0, ∞) and g: R → R be twice differentiable functions such that f'' and g'' are continuous functions on R. Suppose f'(2) = g(2) = 0, f''(2) ≠ 0 and g'(2) ≠ 0. If lim_(x → 2) f(x)g(x)/f'(x)g'(x) = 1, then $f'(2) = 0, \quad g(2) = 0, \quad f''(2) \neq 0, \quad g'(2) \neq 0$ $f > 0; \quad f'', g'' \text{ continuous on } \mathbb{R}$ $\lim_{x\to 2}\frac{f(x)g(x)}{f'(x)g'(x)} = 1$ $\text{Determine which statements hold}$ $f'(2) = 0 \Rightarrow f'(x) = f''(2)(x-2) + o(x-2)$ $g(2) = 0 \Rightarrow g(x) = g'(2)(x-2) + o(x-2)$ $f(x) = f(2) + \tfrac{1}{2}f''(2)(x-2)^2 + o((x-2)^2)$ $\frac{f(x)g(x)}{f'(x)g'(x)} \approx \frac{f(2)\cdot g'(2)(x-2)}{f''(2)(x-2)\cdot g'(2)}$ $= \frac{f(2)}{f''(2)}$ $\lim_{x\to2}\frac{f(x)g(x)}{f'(x)g'(x)} = \frac{f(2)}{f''(2)} = 1$ $\Rightarrow f''(2) = f(2)$ $f(2) > 0 \Rightarrow f''(2) = f(2) > 0$ $f'(2) = 0, \ f''(2) > 0 \Rightarrow \text{local minimum at } x = 2$ $(3): f''(2) > f(2) \text{ is FALSE since } f''(2) = f(2)$ $(4): f(2) - f''(2) = 0 \Rightarrow f(x) - f''(x) = 0 \text{ at } x = 2 \ \checkmark$ $\text{Correct statements: } (1) \text{ and } (4)$