Question

7/ lim \(\sqrt{n^2+4n+1}-n\)

8/ lim \(n-\sqrt{n^2+9n-1}\) (pp liên hợp lim \(\dfrac{n^2-\left(n^2+9n-1\right)}{n+\sqrt{n^2+9n-1}}\)

9/ lim \(\dfrac{1+2+3+...+n}{n^2-1}\)

Answer 1

7/

\(=\lim\dfrac{n^2+4n+1-n^2}{\sqrt{n^2+4n+1}+n}=\lim\dfrac{4n+1}{\sqrt{n^2+4n+1}+n}=\lim\dfrac{4+\dfrac{1}{n}}{\sqrt{1+\dfrac{4}{n}+\dfrac{1}{n^2}}+1}=\dfrac{4}{1+1}=2\)

8/

\(=\lim\dfrac{n^2-\left(n^2+9n-1\right)}{n+\sqrt{n^2+9n-1}}=\lim\dfrac{-9n+1}{n+\sqrt{n^2+9n-1}}=\lim\dfrac{-9+\dfrac{1}{n}}{1+\sqrt{1+\dfrac{9}{n}-\dfrac{1}{n^2}}}=\dfrac{-9}{1+1}=-\dfrac{9}{2}\)

9/

Do \(1+2+...+n=\dfrac{n\left(n+1\right)}{2}=\dfrac{n^2+n}{2}\)

\(\Rightarrow\lim\dfrac{1+2+...+n}{n^2-1}=\lim\dfrac{n^2+n}{2n^2-2}=\lim\dfrac{1+\dfrac{1}{n}}{2-\dfrac{2}{n^2}}=\dfrac{1}{2}\)