Evaluating Limit Product Function and Its Derivatives Properties

To determine the correct statements, we first simplify the expression for $f(x)$ for any $x > 0$.

Given: $f(x) = \lim_{n \to \infty} \left( \frac{n^n(x + n)\left(x + \frac{n}{2}\right) \cdots \left(x + \frac{n}{n}\right)}{n! \left(x^2 + n^2\right)\left(x^2 + \frac{n^2}{4}\right) \cdots \left(x^2 + \frac{n^2}{n^2}\right)} \right)^{\frac{x}{n}}$

Let $P_n(x)$ be the product expression inside the exponent: $P_n(x) = \frac{n^n \prod_{r=1}^n \left(x + \frac{n}{r}\right)}{n! \prod_{r=1}^n \left(x^2 + \frac{n^2}{r^2}\right)}$

Rewriting the product terms in the numerator: $\prod_{r=1}^n \left(x + \frac{n}{r}\right) = \prod_{r=1}^n \frac{n}{r}\left(1 + \frac{rx}{n}\right) = \frac{n^n}{n!} \prod_{r=1}^n \left(1 + \frac{rx}{n}\right)$

Rewriting the product terms in the denominator: $\prod_{r=1}^n \left(x^2 + \frac{n^2}{r^2}\right) = \prod_{r=1}^n \frac{n^2}{r^2}\left(1 + \frac{r^2 x^2}{n^2}\right) = \frac{n^{2n}}{(n!)^2} \prod_{r=1}^n \left(1 + \frac{r^2 x^2}{n^2}\right)$

Substituting these back into $P_n(x)$: $P_n(x) = \frac{n^n \cdot \frac{n^n}{n!} \prod_{r=1}^n \left(1 + \frac{rx}{n}\right)}{n! \cdot \frac{n^{2n}}{(n!)^2} \prod_{r=1}^n \left(1 + \frac{r^2 x^2}{n^2}\right)} = \prod_{r=1}^n \frac{1 + \frac{rx}{n}}{1 + \frac{r^2 x^2}{n^2}}$

Now, taking the natural logarithm of $f(x)$: $\ln f(x) = \ln \left( \lim_{n \to \infty} (P_n(x))^{\frac{x}{n}} \right) = \lim_{n \to \infty} \frac{x}{n} \sum_{r=1}^n \ln \left( \frac{1 + \frac{rx}{n}}{1 + \frac{r^2 x^2}{n^2}} \right)$

Expressing this limit as a definite integral (Riemann sum): $\ln f(x) = x \int_0^1 \ln \left( \frac{1 + tx}{1 + t^2 x^2} \right) dt$

Using the substitution $u = tx$ (where $du = x \, dt$), the limits change from $t=0 \to u=0$ and $t=1 \to u=x$: $\ln f(x) = \int_0^x \left( \ln(1+u) - \ln(1+u^2) \right) du$

Differentiating both sides with respect to $x$ using the Leibniz Rule: $\frac{d}{dx} \big( \ln f(x) \big) = \frac{f'(x)}{f(x)} = \ln(1+x) - \ln(1+x^2) = \ln \left( \frac{1+x}{1+x^2} \right)$

Since $f(x) > 0$ for all $x > 0$:

1. For $x \in (0, 1)$, $\frac{1+x}{1+x^2} > 1 \implies \ln\left(\frac{1+x}{1+x^2}\right) > 0 \implies f'(x) > 0$. Thus, $f(x)$ is strictly increasing on $(0, 1)$.
2. For $x > 1$, $\frac{1+x}{1+x^2} < 1 \implies \ln\left(\frac{1+x}{1+x^2}\right) < 0 \implies f'(x) < 0$. Thus, $f(x)$ is strictly decreasing on $(1, \infty)$.

Now we evaluate each option:

• Option A: Since $f(x)$ is strictly increasing on $(0, 1)$ and $\frac{1}{2} < 1$, we have $f\left(\frac{1}{2}\right) < f(1)$. Thus, option A is incorrect.
• Option B: Since $\frac{1}{3} < \frac{2}{3} < 1$ and $f(x)$ is strictly increasing on $(0, 1)$, $f\left(\frac{1}{3}\right) < f\left(\frac{2}{3}\right)$, which satisfies $f\left(\frac{1}{3}\right) \leq f\left(\frac{2}{3}\right)$. Thus, option B is correct.
• Option C: For $x = 2 > 1$, $f'(2) < 0$, which satisfies $f'(2) \leq 0$. Thus, option C is correct.
• Option D: Calculating $\frac{f'(x)}{f(x)} = \ln \left( \frac{1+x}{1+x^2} \right)$: $\frac{f'(2)}{f(2)} = \ln\left(\frac{3}{5}\right)$ $\frac{f'(3)}{f(3)} = \ln\left(\frac{4}{10}\right) = \ln\left(\frac{2}{5}\right)$ Since $\frac{2}{5} < \frac{3}{5}$, it follows that $\frac{f'(3)}{f(3)} < \frac{f'(2)}{f(2)}$. Thus, option D is incorrect.

Conclusion: The correct statements are B and C.