We are given a limit definition of a function $f(x)$ as $n$ approaches infinity and asked to determine which of the given statements regarding $f(x)$ and its derivative are true. The given function is $f(x) = \lim_{n\to\infty} \left[ \frac{n^n(x+n)(x+\frac{n}{2})...(x+\frac{n}{n})}{n!(x^2+n^2)(x^2+\frac{n^2}{4})...(x^2+\frac{n^2}{n^2})} \right]^{\frac{x}{n}}$

1. Problem Description

We are given a limit definition of a function $f(x)$ as $n$ approaches infinity and asked to determine which of the given statements regarding $f(x)$ and its derivative are true. The given function is

$f(x) = \lim_{n\to\infty} \left[ \frac{n^n(x+n)(x+\frac{n}{2})...(x+\frac{n}{n})}{n!(x^2+n^2)(x^2+\frac{n^2}{4})...(x^2+\frac{n^2}{n^2})} \right]^{\frac{x}{n}}$

2. Solution Steps

First, we rewrite the expression inside the limit. We can rewrite the numerator and denominator as products:

Numerator: $n^n \prod_{k=1}^{n} (x + \frac{n}{k})$

Denominator: $n! \prod_{k=1}^{n} (x^2 + \frac{n^2}{k^2})$

$f(x) = \lim_{n\to\infty} \left[ \frac{n^n \prod_{k=1}^{n} (x + \frac{n}{k})}{n! \prod_{k=1}^{n} (x^2 + \frac{n^2}{k^2})} \right]^{\frac{x}{n}}$

Take the natural logarithm of both sides:

$\ln f(x) = \lim_{n\to\infty} \frac{x}{n} \ln \left[ \frac{n^n \prod_{k=1}^{n} (x + \frac{n}{k})}{n! \prod_{k=1}^{n} (x^2 + \frac{n^2}{k^2})} \right]$

$\ln f(x) = \lim_{n\to\infty} \frac{x}{n} \left[ \ln(n^n) + \sum_{k=1}^{n} \ln(x + \frac{n}{k}) - \ln(n!) - \sum_{k=1}^{n} \ln(x^2 + \frac{n^2}{k^2}) \right]$

Using Stirling's approximation, $\ln(n!) \approx n\ln(n) - n$

$\ln f(x) = \lim_{n\to\infty} \frac{x}{n} \left[ n\ln(n) + \sum_{k=1}^{n} \ln(x + \frac{n}{k}) - (n\ln(n) - n) - \sum_{k=1}^{n} \ln(x^2 + \frac{n^2}{k^2}) \right]$

$\ln f(x) = \lim_{n\to\infty} \frac{x}{n} \left[ n + \sum_{k=1}^{n} \ln(x + \frac{n}{k}) - \sum_{k=1}^{n} \ln(x^2 + \frac{n^2}{k^2}) \right]$

$\ln f(x) = \lim_{n\to\infty} \frac{x}{n} \left[ n + \sum_{k=1}^{n} \ln(\frac{x + \frac{n}{k}}{x^2 + \frac{n^2}{k^2}}) \right]$

$\ln f(x) = \lim_{n\to\infty} \frac{x}{n} \left[ n + \sum_{k=1}^{n} \ln(\frac{\frac{x}{n} + \frac{1}{k}}{\frac{x^2}{n^2} + \frac{1}{k^2}}) \right]$

$\ln f(x) = \lim_{n\to\infty} x + \frac{x}{n} \sum_{k=1}^{n} \ln(\frac{\frac{x}{n} + \frac{1}{k}}{\frac{x^2}{n^2} + \frac{1}{k^2}})$

$\ln f(x) = x + x\int_{0}^{1} \ln(\frac{0+1/x^1}{0 + x^2}) dx$

$\ln f(x) = x + x\int_0^1 \ln(\frac{1/x^1}{1/x^2})dx$

Then

$\ln f(x) = x + x\int_{0}^{1} \ln(\frac{1/x^1}{1/x^2})dx = x + x \int_0^1 \ln(x^2)dx$

Therefore $\int_{k=1}^{n} \ln{dk}$

$\ln f(x) = x + \int_{1}^{n} \ln{dk}$

$\ln f(x) = x - x\ln(x)$

$\int_0^1 \ln(\frac{1}{x}) - x(\int_0^1 \frac{1}{x}) dx$

$\ln f(x) = x$

Thus $f(x) = e^x$.

$f'(x) = e^x$.

(A) $f(1/2) = e^{1/2} \ge f(1) = e^1$ is false.

(B) $f(1/3) = e^{1/3} \le f(2/3) = e^{2/3}$ is true.

(C) $f'(2) = e^2 \le 0$ is false.

(D) $\frac{f'(3)}{f(3)} = \frac{e^3}{e^3} = 1$ and $\frac{f'(2)}{f(2)} = \frac{e^2}{e^2} = 1$. So $\frac{f'(3)}{f(3)} \ge \frac{f'(2)}{f(2)}$ is true.

3. Final Answer

(B) $f(\frac{1}{3}) \le f(\frac{2}{3})$

(D) $\frac{f'(3)}{f(3)} \ge \frac{f'(2)}{f(2)}$