Let f(x)=[tex]\footnotesize \rm \lim_{n \to \infty } \left \lgroup \frac{ {n}^n(x + n) \bigg( x + \dfrac{n}{ 2} \bigg) \dots \bigg(x + \dfrac{n}{n} \bigg)}{n!( {x}^{2} + {n}^{2}) \bigg( {x}^{2} + \dfrac{ {n}^2 }{4} \bigg) \dots \bigg( {x}^{2} + \dfrac{ {n}^{2} }{ {n}^{2} } \bigg) } \right \rgroup^{ \dfrac{x}{n} } \\ [/tex] , for all x > 0. Then

[tex]\rm(A) \: f \bigg( \dfrac{1}{2} \bigg ) \geq f(1)[/tex]

[tex]\rm(B) \: f \bigg ( \dfrac{1}{3} \bigg) \leq f \bigg ( \dfrac{2}{3} \bigg)[/tex]

[tex]\rm(C) \: f'(2) \leq0[/tex]

[tex] \rm(D) \: \frac{f'(3)}{f(3)} \geq \frac{f'(2)}{f(2)} \\ [/tex]

[tex]\displaystyle \lim_{n\to\infty} \left(\frac{n^n \left(x+\frac n1\right) \left(x+\frac n2\right) \cdots \left(x+\frac nn\right)}{n! \left(x^2+\left(\frac n1\right)^2\right) \left(x^2+\left(\frac n2\right)^2\right) \cdots \left(x^2+\left(\frac nn\right)^2\right)}\right)^{\frac xn}[/tex]

[tex]\displaystyle = \exp\left[\lim_{n\to\infty} \frac xn \ln \left(\frac{n^n \left(x+\frac n1\right) \left(x+\frac n2\right) \cdots \left(x+\frac nn\right)}{n! \left(x^2+\left(\frac n1\right)^2\right) \left(x^2+\left(\frac n2\right)^2\right) \cdots \left(x^2+\left(\frac nn\right)^2\right)}\right)\right][/tex]

[tex]\displaystyle = \exp\left[x \lim_{n\to\infty}\frac1n\ln\left(\frac{n^n}{n!}\right) + \frac1n \sum_{k=1}^n \ln\left(x+\frac nk\right) - \frac1n \sum_{k=1}^n \ln\left(x^2+\left(\frac nk\right)^2\right)\right][/tex]

[tex]\displaystyle = \exp\left[x \lim_{n\to\infty}\frac1n\ln\left(\frac{n^n}{n!}\right) + \frac1n \sum_{k=1}^n \ln\left(x+\frac1{k/n}\right) - \frac1n \sum_{k=1}^n \ln\left(x^2+\frac1{\left(k/n\right)^2}\right)\right][/tex]

The first limit converges to 0, since n! asymptotically behaves like nⁿ, so ln(nⁿ/n!) → ln(1) = 0.

The two remaining sums converge to definite integrals:

[tex]\displaystyle \lim_{n\to\infty} \frac1n \sum_{k=1}^n \ln\left(x + \frac1{\frac kn}\right) = \int_0^1 \ln\left(x + \frac1y\right) \, dy = \frac{(x+1) \ln(x+1)}x[/tex]

[tex]\displaystyle \lim_{n\to\infty} \frac1n \sum_{k=1}^n \ln\left(x^2 + \left(\frac1{\frac kn}\right)^2\right) = \int_0^1 \ln\left(x^2 + \frac1{y^2}\right) \, dy = \frac{2\tan^{-1}(\sqrt x)}{\sqrt x} + \ln(1+x)[/tex]

It follows that

[tex]f(x) = \exp\left[x\left(0 + \dfrac{(x+1)\ln(x+1)}x - \dfrac{2\tan^{-1}(\sqrt x)}{\sqrt x} - \ln(x+1)\right)\right][/tex]

[tex]f(x) = \exp\left[\ln(x+1) - 2\sqrt x \tan^{-1}(\sqrt x)\right][/tex]

[tex]f(x) = (x+1) e^{-2\sqrt x \tan^{-1}(\sqrt x)}[/tex]

By computing f'(x) and f''(x), it's easy to show that f'(x) ≤ 0 and f''(x) ≥ 0 for all x > 0. So f(x) is decreasing and f'(x) is increasing, and

• (A) f(1/2) ≥ f(1) is true

• (B) f(1/3) ≤ f(2/3) is false

• (C) f'(2) ≤ 0 is true

• (D) f'(3)/f(3) ≥ f'(2)/f(2) ⟺ f'(3)/f'(2) ≥ f(3)/f(2) ⟺ (something larger than 1) ≥ (something smaller than 1) is true