Question

1) tính giới hạn \(\lim\limits_{n\rightarrow\infty}\left(\sqrt{n^2+6n}-n\right)\)

2) tính giới hạn \(\lim\limits_{n\rightarrow\infty}\left(\sqrt{n+1}-\sqrt{n-1}\right)\)

Answer 1

Lời giải:
1.

\(\lim\limits_{n\to \infty}(\sqrt{n^2+6n}-n)=\lim\limits_{n\to \infty}\frac{6n}{\sqrt{n^2+6n}+n}=\lim\limits_{n\to \infty}\frac{6}{\sqrt{1+\frac{6}{n}}+1}=\frac{6}{1+1}=3\)

2.

\(\lim\limits_{n\to \infty}(\sqrt{n+1}-\sqrt{n-1})=\lim\limits_{n\to \infty}\frac{(n+1)-(n-1)}{\sqrt{n+1}+\sqrt{n-1}}=\lim\limits_{n\to \infty}\frac{2}{\sqrt{n+1}+\sqrt{n-1}}=0\) do $\sqrt{n+1}+\sqrt{n-1}\to \infty$ khi $n\to \infty$