Local Extrema and Derivatives of Functions at a Point

To solve this problem, we will analyze the given limit and apply L'Hôpital's Rule alongside the properties of the functions $f$ and $g$.

Step 1: Evaluate the Limit

We are given the limit: $\lim_{x \to 2} \frac{f(x)g(x)}{f'(x)g'(x)} = 1$

We can rewrite the expression inside the limit as a product of two ratios: $\lim_{x \to 2} \left( \frac{f(x)}{g'(x)} \cdot \frac{g(x)}{f'(x)} \right) = 1$

Given that $f, g$ are twice differentiable and their second derivatives $f'', g''$ are continuous:

1. Since $f: \mathbb{R} \to (0, \infty)$, we have $f(x) > 0$ for all $x \in \mathbb{R}$, which implies $f(2) > 0$. By continuity of $f$, $\lim_{x \to 2} f(x) = f(2) > 0$.
2. Since $g'(2) \neq 0$ and $g''$ is continuous, $g'$ is continuous at $x = 2$, so $\lim_{x \to 2} g'(x) = g'(2) \neq 0$.

Thus, the first factor has a finite non-zero limit: $\lim_{x \to 2} \frac{f(x)}{g'(x)} = \frac{f(2)}{g'(2)}$

Now, consider the second factor $\frac{g(x)}{f'(x)}$ as $x \to 2$: Since $g(2) = 0$ and $f'(2) = 0$, this limit presents an indeterminate form of $\frac{0}{0}$.

Applying L'Hôpital's Rule to $\lim_{x \to 2} \frac{g(x)}{f'(x)}$: $\lim_{x \to 2} \frac{g(x)}{f'(x)} = \lim_{x \to 2} \frac{g'(x)}{f''(x)} = \frac{g'(2)}{f''(2)}$ (This step is valid because $g'$ and $f''$ are continuous at $x = 2$ and $f''(2) \neq 0$).

Step 2: Combine the Limits

Substituting these back into the original limit expression: $\lim_{x \to 2} \frac{f(x)g(x)}{f'(x)g'(x)} = \left( \frac{f(2)}{g'(2)} \right) \cdot \left( \frac{g'(2)}{f''(2)} \right) = \frac{f(2)}{f''(2)}$

Given that this limit equals $1$: $\frac{f(2)}{f''(2)} = 1 \implies f''(2) = f(2)$

Step 3: Analyze the Options

1. Local Extrema at $x = 2$ (Options A and B):

   • We know $f(x) > 0$ for all $x \in \mathbb{R}$, so $f(2) > 0$.
   • Since $f''(2) = f(2)$, it follows that $f''(2) > 0$.
   • We are given $f'(2) = 0$.
   • By the Second Derivative Test, since $f'(2) = 0$ and $f''(2) > 0$, $f$ has a local minimum at $x = 2$.
   • Thus, Option (A) is correct and Option (B) is incorrect.

2. Comparison of $f''(2)$ and $f(2)$ (Option C):

   • From our derivation, $f''(2) = f(2)$, so $f''(2) > f(2)$ is false.
   • Thus, Option (C) is incorrect.

3. Existence of $x \in \mathbb{R}$ such that $f(x) - f''(x) = 0$ (Option D):

   • Since $f''(2) = f(2)$, at $x = 2$ we have: $f(2) - f''(2) = 0$
   • Hence, $f(x) - f''(x) = 0$ holds for at least one $x \in \mathbb{R}$ (namely, $x = 2$).
   • Thus, Option (D) is correct.

Conclusion

The correct options are A and D.