Tính lim[căn bậc hai (1^2+2^2+3^2+...+n^2/2n(n+7)(6n+5)] A.1/6 B.1/2 căn 6 C.1/2 D. dương vô cùng

Ta có: $1^2+2^2+3^2+\mathrm{...}+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$

Khi đó: $\lim \sqrt{\frac{1^2+2^2+3^2+\mathrm{...}+n^2}{2n\left(n+7\right)\left(6n+5\right)}}=\lim \sqrt{\frac{n\left(n+1\right)\left(2n+1\right)}{12n\left(n+7\right)\left(6n+5\right)}}=\lim \sqrt{\frac{\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)}{12\left(1+\frac{7}{n}\right)\left(6+\frac{5}{n}\right)}}=\frac{1}{6}$

Đáp án cần chọn là: A