Đặt

D

- Hướng dẫn giải

Đáp án D

Ta có 

$\begin{array}{l}f\left(n\right)={\left[\left(n^2+1\right)+n\right]}^2+1\\ ={\left(n^2+1\right)}^2+2n\left(n^2+1\right)+n^2+1\end{array}$

$\begin{array}{l}=\left(n^2+1\right)\left[n^2+1+2n+1\right]\\ =\left(n^2+1\right)\left[\left(n^2+1\right)+1\right].\end{array}$ 

Do đó 

$\begin{array}{l}\frac{f\left(2n-1\right)}{f\left(2n\right)}\\ =\frac{\left[{\left(2n-1\right)}^2+1\right].\left[{\left(2n\right)}^2+1\right]}{\left[{\left(2n\right)}^2+1\right].\left[{\left(2n+1\right)}^2+1\right]}\\ =\frac{{\left(2n-1\right)}^2+1}{{\left(2n+1\right)}^2+1}.\end{array}$

Suy ra 

$\begin{array}{l}{u}_n=\frac{f\left(1\right)}{f\left(2\right)}.\frac{f\left(3\right)}{f\left(4\right)}.\frac{f\left(5\right)}{f\left(6\right)}\mathrm{...}\frac{f\left(2n-1\right)}{f\left(2n\right)}\\ =\frac{1^2+1}{3^2+1}.\frac{3^2+1}{5^2+1}.\frac{5^2+1}{7^2+1}\mathrm{...}\frac{{\left(2n-1\right)}^2+1}{{\left(2n+1\right)}^2+1}\end{array}$

$\Rightarrow u_n=\frac{2}{{\left(2n+1\right)}^2}=\frac{1}{2n^2+2n+1}$ 

$\begin{array}{l}\Rightarrow \lim n\sqrt{u_n}=\lim \frac{n}{\sqrt{2n^2+2n+1}}\\ =\frac{1}{\sqrt{2+\frac{1}{n}+\frac{1}{n^2}}}=\frac{1}{\sqrt{2}}.\end{array}$