Mathematics - Limits Question with Solution | TestHub

Taking log both side
$\mathcal{l}n f\left(x\right)= \underset{n\to \infty }{\lim }\frac{x}{n}\mathcal{ }\mathcal{l}n \left[\frac{\prod _{r=1}^n\left(x+\frac{1}{\frac{r}{n}}\right)}{\prod _{r=1}^n\left(x^2+\frac{1}{{\left(\frac{r}{n}\right)}^2}\right) } \frac{1}{\prod _{r=1}^n\left(\frac{r}{n}\right)}\right]$
$=x \underset{n\to \infty }{\lim }\frac{1}{n} \sum _{r=1}^n\mathcal{l}n\left(\frac{x\frac{r}{n}+1}{{\left(x\frac{r}{n}\right)}^2+1}\right)$
Express summation in definite integration using$\frac{r}{n}=\mathrm{t }\&\frac{1}{n}=dt$
$=x \underset{0}{\overset{1}{\int }}\mathcal{l}n\left(\frac{1+tx}{1+t^2{x}^2}\right)dt$
Put$tx=z$
$\mathcal{l}n f\left(x\right)= \underset{0}{\overset{x}{\int }}\mathcal{l}n\left(\frac{1+z}{1+z^2}\right)dz$
$\Rightarrow$$\frac{f\prime \left(x\right)}{f\left(x\right)}= \mathcal{l}n \left(\frac{1+x}{1+x^2}\right)$$\left(\because f\left(x\right)isalwayspositive\right)$
$\forall x\in \left(0,1\right)$
$\ln \frac{1+x}{1+x^2}>0$
$\Rightarrow \frac{f\text{'}\left(x\right)}{f\left(x\right)}>0\forall x\in \left(0,1\right)$
$asf\left(x\right)>0\Rightarrow f\text{'}\left(x\right)>0\forall x\in \left(0,1\right)$
So$f\left(x\right)$is increasing$\left(0,1\right)$and$f^{\prime }\left(1\right)=0$$(x>1\Rightarrow f^{\prime }\left(x\right)<0)$
$\Rightarrow$$f\left(\frac{1}{2}\right)<f\left(1\right), f\left(\frac{1}{3}\right)<f\left(\frac{2}{3}\right), f^{\prime }\left(2\right)<0$
Also$\frac{f^{\prime }\left(3\right)}{f \left(3\right)}-\frac{f^{\prime } \left(2\right)}{f \left(2\right)}=\mathcal{l}n \left(\frac{4}{10}\right)- \mathcal{l}n\left(\frac{3}{5}\right)$
$= \mathcal{l}n \left(\frac{4}{6}\right)<0 \Rightarrow \frac{f^{\prime }\left(3\right)}{f \left(3\right)}<\frac{f^{\prime } \left(2\right)}{f \left(2\right)}$