Question

1) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{6n-8}{n-1}\)

2) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{n^2+5n-3}{4n^3-2n+5}\)

 

Answer 1

1: \(\lim\limits_{n\rightarrow\infty}\dfrac{6n-8}{n-1}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{6-\dfrac{8}{n}}{1-\dfrac{1}{n}}=\dfrac{6-0}{1-0}\)

\(=\dfrac{6}{1}=6\)

2: \(\lim\limits_{n\rightarrow\infty}\dfrac{n^2+5n-3}{4n^3-2n+5}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(1+\dfrac{5}{n}-\dfrac{3}{n^2}\right)}{n^3\left(4-\dfrac{2}{n^2}+\dfrac{5}{n^3}\right)}\)

\(=\lim\limits_{n\rightarrow\infty}\left(\dfrac{1}{n}\cdot\dfrac{1+\dfrac{5}{n}-\dfrac{3}{n^2}}{\left(4-\dfrac{2}{n^2}+\dfrac{5}{n^3}\right)}\right)\)

=0