Question

Let $f(x) = \lim_{n \to \infty} \left( \frac{n^n(x+n)(x+\frac{n}{2})...(x+\frac{n}{n})}{n!(x^2+n^2)(x^2+\frac{n^2}{4})...(x^2+\frac{n^2}{n^2})}\right)^{\frac{x}{n}}$, for all $x>0$. Then

• A. $f(\frac{1}{2}) \ge f(1)$
• B. $f(\frac{1}{3}) \le f(\frac{2}{3})$
• C. $f'(2) \le 0$
• D. $\frac{f'(3)}{f(3)} \ge \frac{f'(2)}{f(2)}$

Answer 1

Answer: (B), (C)

B, C

Answer 2

To determine the correct options, we first need to find an explicit expression for $f(x)$.

Let the expression inside the limit be $A_n$.

$A_n = \frac{n^n(x+n)(x+\frac{n}{2})...(x+\frac{n}{n})}{n!(x^2+n^2)(x^2+\frac{n^2}{4})...(x^2+\frac{n^2}{n^2})}$

We can rewrite the products as follows:

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

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

Substitute these into $A_n$:

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

$A_n = \frac{n^{2n}/n!}{n^{2n}/n!} \frac{\prod_{k=1}^n (1+\frac{xk}{n})}{\prod_{k=1}^n (1+\frac{x^2k^2}{n^2})} = \frac{\prod_{k=1}^n (1+\frac{xk}{n})}{\prod_{k=1}^n (1+\frac{x^2k^2}{n^2})} = \prod_{k=1}^n \frac{1+\frac{xk}{n}}{1+\frac{x^2k^2}{n^2}}$

Now we need to evaluate $f(x) = \lim_{n \to \infty} (A_n)^{\frac{x}{n}}$.

Let $L = f(x)$. Then $\ln L = \lim_{n \to \infty} \frac{x}{n} \ln A_n$.

$\ln A_n = \sum_{k=1}^n \ln\left(\frac{1+\frac{xk}{n}}{1+\frac{x^2k^2}{n^2}}\right)$.

So, $\ln L = \lim_{n \to \infty} \frac{x}{n} \sum_{k=1}^n \ln\left(\frac{1+\frac{xk}{n}}{1+\frac{x^2k^2}{n^2}}\right)$.

This is a Riemann sum. Let $t = k/n$ and $dt = 1/n$.

$\ln L = x \int_0^1 \ln\left(\frac{1+xt}{1+x^2t^2}\right) dt$.

Thus, $f(x) = \exp\left(x \int_0^1 \ln\left(\frac{1+xt}{1+x^2t^2}\right) dt\right)$.

Let $g(x) = \ln f(x) = x \int_0^1 \ln\left(\frac{1+xt}{1+x^2t^2}\right) dt$.

We can evaluate the integral using integration by parts.

$\int_0^1 \ln(1+at) dt = [t\ln(1+at)]_0^1 - \int_0^1 \frac{at}{1+at} dt = \ln(1+a) - \int_0^1 (1-\frac{1}{1+at}) dt$

$= \ln(1+a) - [t - \frac{1}{a}\ln(1+at)]_0^1 = \ln(1+a) - (1 - \frac{1}{a}\ln(1+a)) = (1+\frac{1}{a})\ln(1+a) - 1$.

So, $\int_0^1 \ln(1+xt) dt = (1+\frac{1}{x})\ln(1+x) - 1$.

And $\int_0^1 \ln(1+x^2t^2) dt = (1+\frac{1}{x^2})\ln(1+x^2) - 1$ is incorrect because the coefficient of $t^2$ is $x^2$, not $x$.

For $\int_0^1 \ln(1+x^2t^2) dt$, use $u = \ln(1+x^2t^2)$, $dv = dt$.

$du = \frac{2x^2t}{1+x^2t^2} dt$, $v=t$.

$\int_0^1 \ln(1+x^2t^2) dt = [t\ln(1+x^2t^2)]_0^1 - \int_0^1 \frac{2x^2t^2}{1+x^2t^2} dt$

$= \ln(1+x^2) - \int_0^1 (2 - \frac{2}{1+x^2t^2}) dt = \ln(1+x^2) - [2t - \frac{2}{x}\arctan(xt)]_0^1$

$= \ln(1+x^2) - (2 - \frac{2}{x}\arctan(x))$.

Let $I(x) = \int_0^1 \ln\left(\frac{1+xt}{1+x^2t^2}\right) dt = \int_0^1 \ln(1+xt) dt - \int_0^1 \ln(1+x^2t^2) dt$.

$I(x) = \left((1+\frac{1}{x})\ln(1+x) - 1\right) - \left(\ln(1+x^2) - 2 + \frac{2}{x}\arctan(x)\right)$

$I(x) = (1+\frac{1}{x})\ln(1+x) - \ln(1+x^2) + 1 - \frac{2}{x}\arctan(x)$.

Then $g(x) = x I(x) = x\left((1+\frac{1}{x})\ln(1+x) - \ln(1+x^2) + 1 - \frac{2}{x}\arctan(x)\right)$

$g(x) = (x+1)\ln(1+x) - x\ln(1+x^2) + x - 2\arctan(x)$.

To analyze the options, we need $g'(x) = \frac{f'(x)}{f(x)}$.

$g'(x) = \frac{d}{dx}((x+1)\ln(1+x)) - \frac{d}{dx}(x\ln(1+x^2)) + \frac{d}{dx}(x) - \frac{d}{dx}(2\arctan(x))$

$\frac{d}{dx}((x+1)\ln(1+x)) = \ln(1+x) + (x+1)\frac{1}{1+x} = \ln(1+x) + 1$.

$\frac{d}{dx}(x\ln(1+x^2)) = \ln(1+x^2) + x\frac{2x}{1+x^2} = \ln(1+x^2) + \frac{2x^2}{1+x^2}$.

$\frac{d}{dx}(x) = 1$.

$\frac{d}{dx}(2\arctan(x)) = \frac{2}{1+x^2}$.

$g'(x) = (\ln(1+x) + 1) - (\ln(1+x^2) + \frac{2x^2}{1+x^2}) + 1 - \frac{2}{1+x^2}$

$g'(x) = \ln(1+x) - \ln(1+x^2) + 2 - \frac{2x^2+2}{1+x^2}$

$g'(x) = \ln(1+x) - \ln(1+x^2) + 2 - \frac{2(x^2+1)}{1+x^2}$

$g'(x) = \ln(1+x) - \ln(1+x^2) + 2 - 2$

$g'(x) = \ln\left(\frac{1+x}{1+x^2}\right)$.

Now, let's check the options:

Sign of $g'(x)$:

$g'(x) > 0 \iff \frac{1+x}{1+x^2} > 1 \iff 1+x > 1+x^2 \iff x > x^2 \iff x(1-x) > 0$.

Since $x>0$, this implies $1-x>0 \implies x<1$.

So, $g(x)$ (and $f(x)$) is increasing for $0 < x < 1$.

$g'(x) < 0 \iff \frac{1+x}{1+x^2} < 1 \iff 1+x < 1+x^2 \iff x < x^2 \iff x(1-x) < 0$.

Since $x>0$, this implies $1-x<0 \implies x>1$.

So, $g(x)$ (and $f(x)$) is decreasing for $x > 1$.

$g'(x) = 0 \iff x=1$. This is a local maximum for $f(x)$.

(A) $f(\frac{1}{2}) \ge f(1)$.

Since $1/2 < 1$, and $f(x)$ is increasing on $(0,1)$, $f(1/2) < f(1)$. So (A) is false.

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

Since $1/3 < 2/3$ and both are in $(0,1)$, and $f(x)$ is increasing on $(0,1)$, $f(1/3) < f(2/3)$. So (B) is true.

(C) $f'(2) \le 0$.

Since $2 > 1$, and $f(x)$ is decreasing for $x>1$, $f'(2) < 0$. So (C) is true.

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

This is equivalent to $g'(3) \ge g'(2)$.

To compare $g'(3)$ and $g'(2)$, we need to check the monotonicity of $g'(x)$.

$g''(x) = \frac{d}{dx}\left(\ln(1+x) - \ln(1+x^2)\right) = \frac{1}{1+x} - \frac{2x}{1+x^2} = \frac{1+x^2 - 2x(1+x)}{(1+x)(1+x^2)} = \frac{1+x^2-2x-2x^2}{(1+x)(1+x^2)} = \frac{-x^2-2x+1}{(1+x)(1+x^2)}$.

The denominator is positive for $x>0$. We need to check the sign of the numerator $-x^2-2x+1$.

The roots of $-x^2-2x+1=0$ are $x = \frac{2 \pm \sqrt{4-4(-1)(1)}}{2(-1)} = \frac{2 \pm \sqrt{8}}{-2} = \frac{2 \pm 2\sqrt{2}}{-2} = -1 \mp \sqrt{2}$.

The positive root is $x = -1+\sqrt{2} \approx 0.414$.

Since the parabola $-x^2-2x+1$ opens downwards, it is negative for $x > -1+\sqrt{2}$.

Both $x=2$ and $x=3$ are greater than $-1+\sqrt{2}$.

Therefore, $g''(x) < 0$ for $x > -1+\sqrt{2}$. This means $g'(x)$ is a decreasing function for $x > -1+\sqrt{2}$.

Since $2 < 3$ and $g'(x)$ is decreasing in this interval, $g'(2) > g'(3)$.

So, $\frac{f'(2)}{f(2)} > \frac{f'(3)}{f(3)}$. Thus, option (D) is false.

Both options (B) and (C) are correct.