Đáp án:
$\lim\limits_{n\to \infty}\left(\sqrt2.\sqrt[2^2]{2}\dots\sqrt[2^n]{2}\right)=2$
Giải thích các bước giải:
$\quad \lim\limits_{n\to \infty}\left(\sqrt2.\sqrt[2^2]{2}\dots\sqrt[2^n]{2}\right)$
$=\lim\limits_{n\to \infty}\left(2^{\tfrac12}.2^{\tfrac14}\dots.2^{\tfrac{1}{2^n}}\right)$
$=\lim\limits_{n\to n}2^{\tfrac12+\tfrac14+\cdots+\tfrac{1}{2^n}}$
$=\lim\limits_{n\to \infty}2^{1 -\tfrac{1}{2^n}}$
$=\lim\limits_{n\to \infty}\dfrac{2}{2^{\tfrac{1}{2^n}}}$
$= \dfrac{2}{2^{\lim\limits_{n\to \infty}\tfrac{1}{2^n}}}$
$= \dfrac{2}{2^0}$
$= 2$
