a/ \(=lim\frac{1}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\infty}=0\)

b/ \(=lim\frac{6n+1}{\sqrt{n^2+5n+1}+\sqrt{n^2-n}}=\frac{6+\frac{1}{n}}{\sqrt{1+\frac{5}{n}+\frac{1}{n^2}}+\sqrt{1-\frac{1}{n}}}=\frac{6}{1+1}=3\)

c/ \(=lim\frac{6n-9}{\sqrt{3n^2+2n-1}+\sqrt{3n^2-4n+8}}=lim\frac{6-\frac{9}{n}}{\sqrt{3+\frac{2}{n}-\frac{1}{n^2}}+\sqrt{3-\frac{4}{n}+\frac{8}{n^2}}}=\frac{6}{\sqrt{3}+\sqrt{3}}=\sqrt{3}\)

d/ \(=lim\frac{\left(\frac{2}{6}\right)^n+1-4\left(\frac{4}{6}\right)^n}{\left(\frac{3}{6}\right)^n+6}=\frac{1}{6}\)

e/ \(=lim\frac{\left(\frac{3}{5}\right)^n-\left(\frac{4}{5}\right)^n+1}{\left(\frac{3}{5}\right)^n+\left(\frac{4}{5}\right)^n-1}=\frac{1}{-1}=-1\)

f/ Ta có công thức:

\(1+3+...+\left(2n+1\right)^2=\left(n+1\right)^2\)

\(\Rightarrow lim\frac{1+3+...+2n+1}{3n^2+4}=lim\frac{\left(n+1\right)^2}{3n^2+4}=lim\frac{\left(1+\frac{1}{n}\right)^2}{3+\frac{4}{n^2}}=\frac{1}{3}\)

g/ \(=lim\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\right)=lim\left(1-\frac{1}{n+1}\right)=1-0=1\)

h/ Ta có: \(1^2+2^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)

\(\Rightarrow lim\frac{n\left(n+1\right)\left(2n+1\right)}{6n\left(n+1\right)\left(n+2\right)}=lim\frac{2n+1}{6n+12}=lim\frac{2+\frac{1}{n}}{6+\frac{12}{n}}=\frac{2}{6}=\frac{1}{3}\)

\(=lim\frac{2.2^{5n}+3}{9.3^{5n}+1}=lim\frac{2.\left(\frac{2}{3}\right)^{5n}+3\left(\frac{1}{3}\right)^{5n}}{9+\left(\frac{1}{3}\right)^{5n}}=\frac{0}{9}=0\)

\(b=lim\frac{\left(-\frac{1}{3}\right)^n+4}{-1\left(-\frac{1}{3}\right)^n-2}=\frac{4}{-2}=-2\)

\(c=1+lim\frac{-n}{n^2+\sqrt{n^4+n}}=1+lim\frac{-\frac{1}{n}}{1+\sqrt{1+\frac{1}{n^3}}}=1+\frac{0}{2}=1\)

\(-2\le2cosn^2\le2\Rightarrow\frac{-2}{n^2+1}\le\frac{2cosn^2}{n^2+1}\le\frac{2}{n^2+1}\)

Mà \(lim\frac{-2}{n^2+1}=lim\frac{2}{n^2+1}=0\Rightarrow lim\frac{2cosn^2}{n^2+1}=0\)

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

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

\(=lim\left(\frac{-\frac{2}{n}}{\sqrt{1-\frac{2}{n^2}}+1}-\frac{\frac{2}{n}}{\sqrt[3]{\left(1+\frac{2}{n^2}\right)^2}+\sqrt[3]{1+\frac{2}{n^2}}+1}\right)=\frac{0}{2}-\frac{0}{1+1+1}=0\)

Đặt \(A=\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}\)

Ta có: \(\frac{1}{\sqrt{n^2+1}}>\frac{1}{\sqrt{n^2+n}}>...>\frac{1}{\sqrt{n^2+n}}\)

\(\Rightarrow A< \frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+1}}+...+\frac{1}{\sqrt{n^2+1}}=\frac{n}{\sqrt{n^2+1}}\)

Và \(A>\frac{1}{\sqrt{n^2+n}}+\frac{1}{\sqrt{n^2+n}}+...+\frac{1}{\sqrt{n^2+n}}=\frac{n}{\sqrt{n^2+n}}\)

\(\Rightarrow\frac{n}{\sqrt{n^2+1}}< A< \frac{n}{\sqrt{n^2+n}}\)

Mà \(lim\left(\frac{n}{\sqrt{n^2+1}}\right)=lim\left(\frac{n}{\sqrt{n^2+n}}\right)=1\)

\(\Rightarrow lim\left(A\right)=1\)

Đặt \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{n\left(n+1\right)}=A\)

\(\Leftrightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{n}-\frac{1}{n+1}\)

\(\Leftrightarrow A=\frac{n+1}{n+1}-\frac{1}{n+1}=\frac{n}{n+1}\)

a) lim \(\frac{\left(2n^2-3n+5\right)\left(2n+1\right)}{\left(4-3n\right)\left(2n^2+n+1\right)}\)

= lim \(\frac{\left(2-\frac{3}{n}+\frac{5}{n^2}\right)\left(2+\frac{1}{n}\right)}{\left(\frac{4}{n}-3\right)\left(2+\frac{1}{n}+\frac{1}{n^2}\right)}=\frac{4}{-6}=-\frac{2}{3}\)

b)lim ( \(\frac{\sqrt{n^4+1}}{n}-\frac{\sqrt{4n^6+2}}{n^2}\))

= lim ( \(\frac{n\sqrt{n^4+1}-\sqrt{4n^6+2}}{n^2}\) )

= lim \(\frac{\left(n^6+n^2\right)-\left(4n^6+2\right)}{n^2\left(n\sqrt{n^4+1}+\sqrt{4n^2+2}\right)}\)

= lim \(\frac{-3n^6+n^2+2}{n^3\sqrt{n^4+1}+n^2\sqrt{4n^2+2}}\)

= lim \(\frac{-3n\left(1-\frac{1}{n^4}-\frac{2}{n^6}\right)}{\sqrt{1+\frac{1}{n^4}}+\frac{1}{n^2}\sqrt{4+\frac{2}{n^2}}}\)

= lim \(-3n=-\infty\)

c) lim \(\frac{2n+3}{\sqrt{9n^2+3}-\sqrt[3]{2n^2-8n^3}}\)

= lim\(\frac{2+\frac{3}{n}}{\sqrt{9+\frac{3}{n^2}}-\sqrt[3]{\frac{2}{n}-8}}=\frac{2}{3+2}=\frac{2}{5}\)

a/ \(lim\left(\sqrt[3]{n-n^3}+n+\sqrt{n^2+3n}-n\right)\)

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

\(=lim\left(\frac{1}{\sqrt[3]{n^3+2n+\frac{1}{n}}+\sqrt[3]{n^3-n}+n}+\frac{3}{\sqrt{1+\frac{3}{n}}+1}\right)=0+\frac{3}{1+1}=\frac{3}{2}\)

b/ \(lim\left(\frac{-2\sqrt{n}-4}{\sqrt{n-2\sqrt{n}}+\sqrt{n+4}}\right)=lim\left(\frac{-2-\frac{4}{\sqrt{n}}}{\sqrt{1-\frac{2}{\sqrt{n}}}+\sqrt{1+\frac{4}{n}}}\right)=-\frac{2}{1+1}=-1\)

c/ \(lim\left(\frac{3n^2}{\sqrt[3]{n^6+6n^5+9n^4}+\sqrt[3]{n^6+3n^5}+n^2}\right)=lim\left(\frac{3}{\sqrt[3]{1+\frac{6}{n}+\frac{9}{n^2}}+\sqrt[3]{1+\frac{3}{n}}+1}\right)=\frac{3}{3}=1\)

d/ \(lim\left(\sqrt[3]{n^3+6n}-n+n-\sqrt{n^2-4n}\right)=lim\left(\frac{6n}{\sqrt[3]{n^6+12n^4+36n^2}+\sqrt[3]{n^6+6n^4}+n^2}+\frac{4n}{n+\sqrt{n^2-4n}}\right)\)

\(=lim\left(\frac{6}{\sqrt[3]{n^3+12n+\frac{36}{n}}+\sqrt[3]{n^3+6n}+n}+\frac{4}{1+\sqrt{1-\frac{4}{n}}}\right)=0+\frac{4}{1+1}=2\)

e/ \(lim\left(\frac{-3.3^n+4.4^n}{5.3^n+\frac{3}{2}.4^n}\right)=lim\left(\frac{-3\left(\frac{3}{4}\right)^n+4}{5.\left(\frac{3}{4}\right)^n+\frac{3}{2}}\right)=\frac{0+4}{0+\frac{3}{2}}=\frac{8}{3}\)

f/ \(lim\left(\frac{9^n-5.5^n+7.7^n}{9.3^n+5^n+2.8^n}\right)=lim\left(\frac{1-5.\left(\frac{5}{9}\right)^n+7\left(\frac{7}{9}\right)^n}{9.\left(\frac{1}{3}\right)^n+\left(\frac{5}{9}\right)^n+2.\left(\frac{8}{9}\right)^n}\right)=\frac{1}{0}=+\infty\)

g/ \(lim\left(\frac{6.6^n+3^5.9^n}{3^3.9^n-\frac{1}{2}.4^n}\right)=lim\left(\frac{6\left(\frac{2}{3}\right)^n+3^5}{3^3-\frac{1}{2}\left(\frac{4}{9}\right)^n}\right)=\frac{3^5}{3^3}=9\)

\(a=\lim\limits n\left(\sqrt[3]{\frac{1}{n}+1}+1\right)=+\infty.2=+\infty\)

\(b=\lim\limits\frac{n^2+2\sqrt{n}+3}{2n^2+n-\sqrt{n}}=\lim\limits\frac{1+\frac{2}{n\sqrt{n}}+\frac{3}{n^2}}{2+\frac{1}{n}-\frac{1}{n\sqrt{n}}}=\frac{1}{2}\)

\(c=\lim\limits\frac{2n\sqrt{n}+3}{n^2+n+1}=\frac{\frac{2}{\sqrt{n}}+\frac{3}{n^2}}{1+\frac{1}{n}+\frac{1}{n^2}}=\frac{0}{1}=0\)

\(d=\lim\limits\frac{2n^2+6n\sqrt{n}}{n^2+3n+2}=\lim\limits\frac{2+\frac{6}{\sqrt{n}}}{1+\frac{3}{n}+\frac{2}{n^2}}=\frac{2}{1}=2\)

a) \(\lim \frac{{5n + 1}}{{2n}} = \lim \frac{{5 + \frac{1}{n}}}{2} = \frac{{5 + 0}}{2} = \frac{5}{2}\)           

b) \(\lim \frac{{6{n^2} + 8n + 1}}{{5{n^2} + 3}} = \lim \frac{{6 + \frac{8}{n} + \frac{1}{{{n^2}}}}}{{5 + \frac{3}{{{n^2}}}}} = \frac{{6 + 0 + 0}}{{5 + 0}} = \frac{6}{5}\)                   

c) \(\lim \frac{{\sqrt {{n^2} + 5n + 3} }}{{6n + 2}} = \lim \frac{{\sqrt {1 + \frac{5}{n} + \frac{3}{{{n^2}}}} }}{{6 + \frac{2}{n}}} = \frac{{\sqrt {1 + 0 + 0} }}{{6 + 0}} = \frac{1}{6}\)

d) \(\lim \left( {2 - \frac{1}{{{3^n}}}} \right) = \lim 2 - \lim {\left( {\frac{1}{3}} \right)^n} = 2 - 0 = 0\)              

e) \(\lim \frac{{{3^n} + {2^n}}}{{{{4.3}^n}}} = \lim \frac{{1 + {{\left( {\frac{2}{3}} \right)}^n}}}{4} = \frac{{1 + 0}}{4} = \frac{1}{4}\)                       

g) \(\lim \frac{{2 + \frac{1}{n}}}{{{3^n}}}\)

Ta có \(\lim \left( {2 + \frac{1}{n}} \right) = \lim 2 + \lim \frac{1}{n} = 2 + 0 = 2 > 0;\lim {3^n} =  + \infty  \Rightarrow \lim \frac{{2 + \frac{1}{n}}}{{{3^n}}} = 0\)