Tính \(\lim\dfrac{\left(2n+1\right)\left(n^2+3n\right)}{\left(n+1\right)^2.n}\) là
2.3.1.4.Hướng dẫn giải:Có \(\dfrac{\left(2n+1\right)\left(n^2+3n\right)}{\left(n+1\right)^2.n}\text{}\)\(=\dfrac{\left(2-\dfrac{1}{n}\right)\left(1+\dfrac{3}{n}\right)}{\left(1+\dfrac{1}{n}\right)^2.1}\) nên \(\lim\dfrac{\left(2n+1\right)\left(n^2+3n\right)}{\left(n+1\right)^2.n}=\dfrac{\left(2-0\right)\left(1+0\right)}{\left(1+0\right)^2}=2.\)
