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

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

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

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

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

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

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

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

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

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

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

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

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

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: 

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

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

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

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

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

\(=0\)

3:

\(\lim\limits_{n\rightarrow\infty}\dfrac{2-5^{n-2}}{3^n+2\cdot5^n}\)

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

\(=\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{2}{5^n}-\dfrac{1}{25}}{\left(\dfrac{3}{5}\right)^n+2\cdot1}\)

\(=-\dfrac{1}{25}:2=-\dfrac{1}{50}\)

1:

\(=\lim\limits_{n\rightarrow\infty}\dfrac{3^n-4^n\cdot4}{3^n\cdot9+4^n}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{3^n}{4^n}-4}{3^n\cdot\dfrac{9}{4^n}+1}\)

\(=-\dfrac{4}{1}=-4\)

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

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

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

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

\(=+\infty\)

\(a=\lim\dfrac{1}{\sqrt{4n+1}+2\sqrt{n}}=\dfrac{1}{\infty}=0\)

\(b=\lim n\left(\sqrt{1+\dfrac{2}{n}}-\sqrt{1-\dfrac{2}{n}}-1\right)=+\infty.\left(-1\right)=-\infty\)

\(c=\lim4^n\left(\sqrt{\left(\dfrac{9}{16}\right)^n-\left(\dfrac{3}{16}\right)^n}-1\right)=+\infty.\left(-1\right)=-\infty\)

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

\(a=\lim4^n\left(1-\left(\dfrac{3}{4}\right)^n\right)=+\infty.1=+\infty\)

\(b=\lim\left(4^n+2.2^n+1-4^n\right)=\lim2^n\left(2+\dfrac{1}{2^n}\right)=+\infty.2=+\infty\)

\(c=limn^3\left(\sqrt{\dfrac{2}{n}-\dfrac{3}{n^4}+\dfrac{11}{n^6}}-1\right)=+\infty.\left(-1\right)=-\infty\)

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

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

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

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