Bạn xem lại đề ( $n$ tiến đến đâu?)

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

\(a,lim\dfrac{^3\sqrt{8n^3+2n}}{-n+3}\)

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

\(\lim\dfrac{\left(2n\sqrt{n}+1\right)\left(\sqrt{n}+3\right)}{\left(n-1\right)\left(3-2n\right)}=\lim\dfrac{\left(2+\dfrac{1}{n\sqrt{n}}\right)\left(1+\dfrac{3}{\sqrt{n}}\right)}{\left(1-\dfrac{1}{n}\right)\left(\dfrac{3}{n}-2\right)}=\dfrac{2.1}{1.\left(-2\right)}=-1\)

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

\(\lim\limits\left(2-3n\right)^4\left(n+1\right)^3=\lim n^7\left(3-\dfrac{2}{n}\right)^4\left(1+\dfrac{1}{n}\right)^3=+\infty\)

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

\(\lim\left(\sqrt[3]{8n^3+3n^2+4}-2n+6\right)=\lim\dfrac{8n^3+3n^2+4-\left(2n-6\right)^3}{\sqrt[3]{\left(8n^3+3n^2+4\right)^2}+\left(2n-6\right)\sqrt[3]{8n^3+3n^2+4}+\left(2n-6\right)^2}\)

\(=\lim\dfrac{75n^2-216n+220}{\sqrt[3]{\left(8n^3+3n^2+4\right)^2}+\left(2n-6\right)\sqrt[3]{8n^3+3n^2+4}+\left(2n-6\right)^2}\)

\(=\lim\dfrac{75-\dfrac{216}{n}+\dfrac{220}{n^2}}{\sqrt[3]{\left(8+\dfrac{3}{n}+\dfrac{4}{n^3}\right)^2}+\left(2-\dfrac{6}{n}\right)\sqrt[3]{8+\dfrac{3}{n}+\dfrac{4}{n^3}}+\left(2-\dfrac{6}{n}\right)^2}\)

\(=\dfrac{75}{\sqrt[3]{8^2}+2.\sqrt[3]{8}+2^2}=...\)

d.

\(\lim\left(\sqrt[3]{8n^3+3n^2-2}+\sqrt[3]{5n^2-8n^3}\right)\)

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

\(=\lim\dfrac{8n^3+3n^2-2-\left(8n^3-5n^2\right)}{\sqrt[3]{\left(8n^3+3n^2-2\right)^2}+\sqrt[3]{\left(8n^3+3n^2-2\right)\left(8n^3-5n^2\right)}+\sqrt[3]{8n^3-5n^2}}\)

\(=\lim\dfrac{8n^2-2}{\sqrt[3]{\left(8n^3+3n^2-2\right)^2}+\sqrt[3]{\left(8n^3+3n^2-2\right)\left(8n^3-5n^2\right)}+\sqrt[3]{8n^3-5n^2}}\)

\(=lim\dfrac{8-\dfrac{2}{n^2}}{\sqrt[3]{\left(8+\dfrac{3}{n}-\dfrac{2}{n^3}\right)^2}+\sqrt[3]{\left(8+\dfrac{3}{n}-\dfrac{2}{n^3}\right)\left(8-\dfrac{5}{n}\right)}+\sqrt[3]{\left(8-\dfrac{5}{n}\right)^2}}\)

\(=\dfrac{8}{\sqrt[3]{8^2}+\sqrt[3]{8.8}+\sqrt[3]{8^2}}=...\)

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

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

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

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

\(=\dfrac{2}{\sqrt[]{4}+2}+\dfrac{3}{4+2\sqrt[3]{8}+\sqrt[3]{8^2}}=...\)