Đáp án: $ \lim n\sqrt{u_n}=\dfrac{1}{\sqrt2}$
Giải thích các bước giải:
Ta có:
$\dfrac{f(2n-1)}{f(2n)}=\dfrac{(4n^2-2n+1)^2+1}{(4n^2+2n+1)^2+1}$
$\to \dfrac{f(2n-1)}{f(2n)}=\dfrac{(4n^2+1-2n)^2+1}{(4n^2+1+2n)^2+1}$
$\to \dfrac{f(2n-1)}{f(2n)}=\dfrac{(4n^2+1)^2-4n(4n^2+1)+4n^2+1}{(4n^2+1)^2+4n(4n^2+1)+4n^2+1}$
$\to \dfrac{f(2n-1)}{f(2n)}=\dfrac{(4n^2+1)(4n^2+1-4n+1)}{(4n^2+1)(4n^2+1+4n+1)}$
$\to \dfrac{f(2n-1)}{f(2n)}=\dfrac{4n^2+1-4n+1}{4n^2+1+4n+1}$
$\to \dfrac{f(2n-1)}{f(2n)}=\dfrac{(2n-1)^2+1}{(2n+1)^2+1}$
$\to u_n=\dfrac{f(1)}{f(2)}\cdot \dfrac{f(3)}{f(4)}\cdots\dfrac{f(2n-1)}{f(2n)}$
$\to u_n=\dfrac2{10}\cdot \dfrac{10}{26}\cdots\dfrac{(2n-1)^2+1}{(2n+1)^2+1}$
$\to u_n=\dfrac{2}{(2n+1)^2+1}$
$\to \lim n\sqrt{u_n}=\lim \sqrt{\dfrac{2n^2}{4n^2+4n+2}}$
$\to \lim n\sqrt{u_n}=\lim \sqrt{\dfrac{2}{4+\dfrac4n+\dfrac2{n^2}}}$
$\to \lim n\sqrt{u_n}=\dfrac{1}{\sqrt2}$
