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Lời giải:
Xét hạng tử tổng quát:
\(\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}=\frac{(n+1)-n}{\sqrt{n(n+1)}(\sqrt{n}+\sqrt{n+1)}}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}\)
\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Cho $n=1,2,...$ thì:
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}=1-\frac{1}{\sqrt{2}}\)
\(\frac{1}{3\sqrt{2}+2\sqrt{3}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)
......
\(\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
\(\Rightarrow U_n=1-\frac{1}{\sqrt{n+1}}\)
\(\Rightarrow \lim\limits U_n=\lim (1-\frac{1}{\sqrt{n+1}})=1\)
\(\frac{1}{\sqrt[3]{2}}>\frac{1}{\sqrt[3]{3}}>...>\frac{1}{\sqrt[3]{n}}\)
\(\Rightarrow\frac{n-1}{\sqrt[3]{n}}< f\left(n\right)< \frac{n-1}{\sqrt[3]{2}}\)
Mà \(\lim\limits\frac{n-1}{\sqrt[3]{n}\left(n^2+1\right)}=\lim\limits\frac{n-1}{\sqrt[3]{n}\left(n^2+1\right)}=0\)
\(\Rightarrow\lim\limits\frac{f\left(n\right)}{n^2+1}=0\)
\(a=\lim\limits n\left(\sqrt[3]{\frac{1}{n}+1}+1\right)=+\infty.2=+\infty\)
\(b=\lim\limits\frac{n^2+2\sqrt{n}+3}{2n^2+n-\sqrt{n}}=\lim\limits\frac{1+\frac{2}{n\sqrt{n}}+\frac{3}{n^2}}{2+\frac{1}{n}-\frac{1}{n\sqrt{n}}}=\frac{1}{2}\)
\(c=\lim\limits\frac{2n\sqrt{n}+3}{n^2+n+1}=\frac{\frac{2}{\sqrt{n}}+\frac{3}{n^2}}{1+\frac{1}{n}+\frac{1}{n^2}}=\frac{0}{1}=0\)
\(d=\lim\limits\frac{2n^2+6n\sqrt{n}}{n^2+3n+2}=\lim\limits\frac{2+\frac{6}{\sqrt{n}}}{1+\frac{3}{n}+\frac{2}{n^2}}=\frac{2}{1}=2\)
a/ \(lim\left(\sqrt[3]{n-n^3}+n+\sqrt{n^2+3n}-n\right)\)
\(=lim\left(\frac{n}{\sqrt[3]{\left(n-n^3\right)^2}-n\sqrt[3]{\left(n-n^3\right)}+n^2}+\frac{3n}{\sqrt{n^2+3n}+n}\right)\)
\(=lim\left(\frac{1}{\sqrt[3]{n^3+2n+\frac{1}{n}}+\sqrt[3]{n^3-n}+n}+\frac{3}{\sqrt{1+\frac{3}{n}}+1}\right)=0+\frac{3}{1+1}=\frac{3}{2}\)
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 \(\frac{\left(2n^2-3n+5\right)\left(2n+1\right)}{\left(4-3n\right)\left(2n^2+n+1\right)}\)
= 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/ \(=lim\frac{1}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\infty}=0\)
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: \(1^2+2^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)
\(\Rightarrow lim\frac{n\left(n+1\right)\left(2n+1\right)}{6n\left(n+1\right)\left(n+2\right)}=lim\frac{2n+1}{6n+12}=lim\frac{2+\frac{1}{n}}{6+\frac{12}{n}}=\frac{2}{6}=\frac{1}{3}\)
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 \(lim\left(\frac{n}{\sqrt{n^2+1}}\right)=lim\left(\frac{n}{\sqrt{n^2+n}}\right)=1\)
\(\Rightarrow lim\left(A\right)=1\)
\(\frac{1}{\sqrt{n^2+1}}=\frac{1}{\sqrt{n^2+1}}\)
\(\frac{1}{\sqrt{n^2+2}}< \frac{1}{\sqrt{n^2+1}}\)
...
\(\frac{1}{\sqrt{n^2+n}}< \frac{1}{\sqrt{n^2+1}}\)
Cộng lại ta được: \(A< \frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+1}}+...+\frac{1}{\sqrt{n^2+1}}\)
Tương tự:
\(\frac{1}{\sqrt{n^2+1}}>\frac{1}{\sqrt{n^2+n}}\)
\(\frac{1}{\sqrt{n^2+2}}>\frac{1}{\sqrt{n^2+n}}\)
...
\(\frac{1}{\sqrt{n^2+n}}=\frac{1}{\sqrt{n^2+n}}\)
Cộng lại ta được:
\(A>\frac{1}{\sqrt{n^2+n}}+\frac{1}{\sqrt{n^2+n}}+...+\frac{1}{\sqrt{n^2+n}}\)
\(\frac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
\(\Rightarrow S_n=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
\(\Rightarrow S_n=1-\frac{1}{\sqrt{n+1}}\)
\(lim\left(S_n\right)=lim\left(1-\frac{1}{\sqrt{n+1}}\right)=1-0=1\)