Question

  Giúp em với ạaa.

Answer 1

1: \(\lim\limits_{n\rightarrow+\infty}\left(\sqrt{n^2+n+1}+\sqrt{4n^2+2n+1}-3n\right)\)

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

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

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

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

\(=\dfrac{1+0}{1+1}+\dfrac{2}{2+2}=\dfrac{1}{2}+\dfrac{2}{4}=1\)