Đáp án: $\dfrac23$
Giải thích các bước giải:
Ta có:
$\dfrac{n^3-1}{n^3+1}=\dfrac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}$
$\to \dfrac{n^3-1}{n^3+1}=\dfrac{(n-1)((n+1)^2-(n+1)+1)}{(n+1)(n^2-n+1)}$
Ta có:
$L=\lim(\dfrac{2^3-1}{2^3+1}\cdot \dfrac{3^3-1}{3^3+1}\cdot \dfrac{4^3-1}{4^3+1}\cdots\dfrac{n^3-1}{n^3+1})$
$\to L=\lim \dfrac{(2-1)((2+1)^2-(2+1)+1)}{(2+1)(2^2-2+1)}\cdot \dfrac{(3-1)((3+1)^2-(3+1)+1)}{(3+1)(3^2-3+1)}\cdot \dfrac{(4-1)((4+1)^2-(4+1)+1)}{(4+1)(4^2-4+1)}\cdots\dfrac{(n-1)((n+1)^2-(n+1)+1)}{(n+1)(n^2-n+1)}$
$\to L=\lim \dfrac{1\cdot (3^2-3+1)}{3\cdot (2^2-2+1)}\cdot \dfrac{2\cdot (4^2-4+1)}{4\cdot (3^2-3+1)}\cdot \dfrac{3\cdot (5^2-5+1)}{5\cdot (4^2-4+1)}\cdots\dfrac{(n-1)((n+1)^2-(n+1)+1)}{(n+1)(n^2-n+1)}$
$\to L=\lim\dfrac{2\cdot ((n+1)^2-(n+1)+1)}{(2^2-2+1)\cdot n\cdot (n+1)}$
$\to L=\dfrac{2(n^2+n+1)}{3n(n+1)}$
$\to L=\dfrac{2(1+\dfrac1n+\dfrac1{n^2})}{3(1+\dfrac1n)}$
$\to L=\dfrac23$
