Question

1) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{n^2-n+2}{n^3+2n^2-3}\)

2) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{n+2}{3n^3-2n+n^2}\)

Answer 1

1: \(\lim\limits_{n\rightarrow\infty}\dfrac{n^2-n+2}{n^3+2n^2-3}=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(1-\dfrac{1}{n}+\dfrac{2}{n^2}\right)}{n^3\left(1+\dfrac{2}{n}-\dfrac{3}{n^3}\right)}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{1-\dfrac{1}{n}+\dfrac{2}{n^2}}{n\left(1+\dfrac{2}{n}-\dfrac{3}{n^3}\right)}=\lim\limits_{n\rightarrow\infty}\dfrac{1}{n}=0\)

2: 

\(\lim\limits_{n\rightarrow\infty}\dfrac{n+2}{3n^3+n^2-2n}=\lim\limits_{n\rightarrow\infty}\dfrac{n\left(1+\dfrac{2}{n}\right)}{n^3\left(3+\dfrac{1}{n}-\dfrac{2}{n^2}\right)}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{1}{n^2}=0\)