Tính giới hạn $\large L = \lim\dfrac{(n^2+2n)(2n^3+1)(4n+5)}{(n^4-3n-1

Tính giới hạn $\large L = \lim\dfrac{(n^2+2n)(2n^3+1)(4n+5)}{(n^4-3n-1)(3n^2-7)}$

• A. $\large L=0$
• B. $\large L = 1$
• C. $\large L = \dfrac{8}{3}$
• D. $\large L = +\infty$

Chọn C

Ta có: $\large L=\lim\dfrac{(n^2+2n)(2n^3+1)(4n+5)}{(n^4-3n-1)(3n^2-7)}=\lim\dfrac{\left(1+\dfrac{2}{n} \right )\left(2+\dfrac{1}{n^3} \right )\left(4+\dfrac{5}{n} \right )}{\left(1-\dfrac{3}{n^3}-\dfrac{1}{n^4} \right )\left(3-\dfrac{7}{n^2} \right )}=\dfrac{1.2.4}{1.3}=\dfrac{8}{3}$