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

\(\dfrac{-n^2+2n+1}{\sqrt{3n^4+2}}=\dfrac{-\dfrac{n^2}{n^4}+\dfrac{2n}{n^4}+\dfrac{1}{n^4}}{\sqrt{\dfrac{3n^4}{n^4}+\dfrac{2}{n^4}}}=\dfrac{-\dfrac{1}{n^2}+\dfrac{2}{n^3}+\dfrac{1}{n^4}}{\sqrt{3+\dfrac{2}{n^4}}}\)

\(\Rightarrow\left(1\right)=\dfrac{-lim\dfrac{1}{n^2}+2lim\dfrac{1}{n^3}+lim\dfrac{1}{n^4}}{\sqrt{lim\left(3+\dfrac{2}{n^4}\right)}}\)

\(=\dfrac{0}{\sqrt{lim\left(3+\dfrac{2}{n^4}\right)}}=0\)

\(2,\lim\limits_{n\rightarrow\infty}\left(\dfrac{4n-\sqrt{16n^2+1}}{n+1}\right)\left(2\right)\)

\(\dfrac{4n-\sqrt{16n^2+1}}{n+1}=\dfrac{\dfrac{4n}{n^2}-\sqrt{\dfrac{16n^2}{n^2}+\dfrac{1}{n^2}}}{\dfrac{n}{n^2}+\dfrac{1}{n^2}}=\dfrac{\dfrac{4}{n}-\sqrt{16+\dfrac{1}{n^2}}}{\dfrac{1}{n}+\dfrac{1}{n^2}}\)

\(\Rightarrow\left(2\right)=\dfrac{lim\left(\dfrac{4}{n}-\sqrt{16+\dfrac{1}{n^2}}\right)}{lim\left(\dfrac{1}{n}+\dfrac{1}{n^2}\right)}=\dfrac{lim\left(\dfrac{4}{n}-\sqrt{16+\dfrac{1}{n^2}}\right)}{0}\)

Vậy giới hạn \(\left(2\right)\) không xác định.

\(3,\lim\limits_{n\rightarrow\infty}\left(\dfrac{\sqrt{9n^2+n+1}-3n}{2n}\right)\left(3\right)\)

\(\dfrac{\sqrt{9n^2+n+1}-3n}{2n}=\dfrac{\sqrt{9+\dfrac{1}{n}+\dfrac{1}{n^2}}-\dfrac{3}{n}}{\dfrac{2}{n}}\)

\(\Rightarrow\left(3\right)=\dfrac{lim\left(\sqrt{9+\dfrac{1}{n}+\dfrac{1}{n^2}}-\dfrac{3}{n}\right)}{2lim\dfrac{1}{n}}=\dfrac{lim\left(\sqrt{9+\dfrac{1}{n}+\dfrac{1}{n^2}}-\dfrac{3}{n}\right)}{0}\)

Vậy \(lim\left(3\right)\) không xác định.

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

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

\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2+n+1}{n^2\cdot\sqrt[3]{\left(1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}\right)^2}+n^2\cdot\sqrt[3]{1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}}+n^2}\)

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

\(=\dfrac{1}{1+1+1}=\dfrac{1}{3}\)

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

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

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

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

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

1:

\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^3+3n^2+1-n^3}{\sqrt[3]{n^3+3n^2+1}+n}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{3n^2+1}{\sqrt[3]{n^3+3n^2+1}+n}\)

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

\(=\lim\limits_{n\rightarrow\infty}\dfrac{n\cdot\left(3+\dfrac{1}{n^2}\right)}{\sqrt[3]{1+\dfrac{3}{n}+\dfrac{1}{n^3}}+1}\)

\(=\lim\limits_{n\rightarrow\infty}n\cdot\lim\limits_{n\rightarrow\infty}\dfrac{3+\dfrac{1}{n^2}}{\sqrt[3]{1+\dfrac{3}{n}+\dfrac{1}{n^3}}+1}\)

\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{n\rightarrow\infty}n=+\infty\\\lim\limits_{n\rightarrow\infty}\dfrac{3+\dfrac{1}{n^2}}{\sqrt[3]{1+\dfrac{3}{n}+\dfrac{1}{n^3}}+1}=\dfrac{3}{2}>0\end{matrix}\right.\)

2: 

\(=\lim\limits_{n\rightarrow\infty}\left(\sqrt{4n^2+1}-2n+2n-\sqrt[3]{8n^3+n}\right)\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{4n^2+1-4n^2}{\sqrt{4n^2+1}+2n}+\lim\limits_{n\rightarrow\infty}\dfrac{8n^3-8n^3-n}{4n^2+2n\cdot\sqrt[3]{8n^3+n}+\left(\sqrt[3]{8n^3+n}\right)^2}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{1}{\sqrt{4n^2+1}+2n}+\lim\limits_{n\rightarrow\infty}\dfrac{-n}{4n^2+2n\cdot n\cdot\sqrt[3]{8+\dfrac{1}{n^3}}+\left(n\cdot\sqrt[3]{8+\dfrac{1}{n^3}}\right)^2}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{-n}{4n^2+2n^2\cdot\sqrt[3]{8+\dfrac{1}{n^3}}+n^2\cdot\sqrt[3]{\left(8+\dfrac{1}{n^3}\right)^2}}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{-1}{4n+2n\cdot\sqrt[3]{8+\dfrac{1}{n^3}}+n\cdot\sqrt[3]{\left(8+\dfrac{1}{n^3}\right)^2}}\)

\(=0\)

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 $n\to \infty$

1:

\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2-1-9n^2}{\sqrt{n^2-1}-3n}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{-8n^2-1}{\sqrt{n^2-1}-3n}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(-8-\dfrac{1}{n^2}\right)}{n\left(\sqrt{1-\dfrac{1}{n^2}}-3\right)}=\lim\limits_{n\rightarrow\infty}-\dfrac{8}{1-3}\cdot n=\lim\limits_{n\rightarrow\infty}4n=+\infty\)

2: 

\(\lim\limits_{n\rightarrow\infty}\sqrt{4n^2+5}+n\)

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

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

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

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

1: \(I=\lim\limits_{n\rightarrow\infty}\dfrac{n^2+2-n^2+1}{\sqrt{n^2+2}+\sqrt{n^2-1}}\)

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

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

=0

2: \(\lim\limits_{n\rightarrow\infty}\sqrt{n^2+2n+2}+n\)

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

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

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

\(=+\infty\)

1. \(lim_{n\rightarrow+\infty}\dfrac{n^2+1}{n}=lim_{n\rightarrow+\infty}\left(n+\dfrac{1}{n}\right)=+\infty\)(đpcm)

2. \(lim_{n\rightarrow+\infty}\dfrac{2-n}{\sqrt{n}}=lim_{n\rightarrow+\infty}\left(2-\sqrt{n}\right)=-\infty\) (đpcm)

a) \(\lim\limits\dfrac{2n^2+3n}{n^2+1}=\lim\limits\dfrac{n^2\left(2+\dfrac{3n}{n^2}\right)}{n^2\left(1+\dfrac{1}{n^2}\right)}=\lim\limits\dfrac{2+\dfrac{3}{n}}{1+\dfrac{1}{n^2}}=2\).

b) \(\lim\limits\dfrac{\sqrt{4n^2+3}}{n}\\ =\lim\limits\dfrac{\sqrt{n^2\left(4+\dfrac{3}{n^2}\right)}}{n}\\ =\lim\limits\dfrac{\sqrt[n]{4+\dfrac{3}{n^2}}}{n}\\ =\lim\limits\sqrt{4+\dfrac{3}{n^2}}\\ =2.\)

a) \(\lim \frac{{ - 2n + 1}}{n} = \lim \frac{{n\left( { - 2 + \frac{1}{n}} \right)}}{n} = \lim \left( { - 2 + \frac{1}{n}} \right) =  - 2\)

b) \(\lim \frac{{\sqrt {16{n^2} - 2} }}{n} = \lim \frac{{\sqrt {{n^2}\left( {16 - \frac{2}{{{n^2}}}} \right)} }}{n} = \lim \frac{{n\sqrt {16 - \frac{2}{{{n^2}}}} }}{n} = \lim \sqrt {16 - \frac{2}{{{n^2}}}}  = 4\)

c) \(\lim \frac{4}{{2n + 1}} = \lim \frac{4}{{n\left( {2 + \frac{1}{n}} \right)}} = \lim \left( {\frac{4}{n}.\frac{1}{{2 + \frac{1}{n}}}} \right) = \lim \frac{4}{n}.\lim \frac{1}{{2 + \frac{1}{n}}} = 0\)

d) \(\lim \frac{{{n^2} - 2n + 3}}{{2{n^2}}} = \lim \frac{{{n^2}\left( {1 - \frac{2}{n} + \frac{3}{{{n^2}}}} \right)}}{{2{n^2}}} = \lim \frac{{1 - \frac{2}{n} + \frac{3}{{{n^2}}}}}{2} = \frac{1}{2}\)