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1.
\(lim\frac{3n+1}{\sqrt[3]{\left(n^3+3n+1\right)^2}+n\sqrt{n^3+3n+1}+n^2}=lim\frac{3+\frac{1}{n}}{\sqrt[3]{\frac{\left(n^3+3n+1\right)^2}{n^3}}+\sqrt{n^3+3n+1}+n}=\frac{3}{\infty}=0\)
b=\(lim\left(\sqrt[3]{n^3+2n}-n+n-\sqrt{n^2+1}\right)=lim\left(\frac{2n}{\sqrt[3]{\left(n^3+2n\right)^2}+n\sqrt[3]{n^3+2n}+n^2}-\frac{1}{n+\sqrt{n^2+1}}\right)\)
\(=lim\left(\frac{2}{\sqrt[3]{\frac{\left(n^3+2n\right)^2}{n^3}}+\sqrt[3]{n^3+2n}+n}-\frac{1}{n+\sqrt{n^2+1}}\right)=0-0=0\)
c\(=lim\left(\frac{2n^2+n}{\sqrt[3]{\left(n^3+n\right)^2}+\sqrt[3]{\left(n^3+n\right)\left(n^3-2n^2\right)}+\sqrt[3]{\left(n^3-2n^2\right)^2}}\right)\)
\(=lim\left(\frac{2+\frac{1}{n}}{\sqrt[3]{\left(1+\frac{1}{n^2}\right)^2}+\sqrt[3]{\left(1+\frac{1}{n^2}\right)\left(1-\frac{2}{n}\right)}+\sqrt[3]{\left(1-\frac{2}{n}\right)^2}}\right)=\frac{2}{1+1.1+1}=\frac{2}{3}\)
2.
a\(=lim\left[n\left(2-\sqrt{1+\frac{3}{n}}\right)\right]=+\infty\left(2-1\right)=+\infty\)
\(b=lim\left[n\left(\sqrt{1+\frac{2}{n^2}}-\sqrt{\frac{3}{n}+\frac{1}{n^2}}\right)\right]=+\infty\left(1-0\right)=+\infty\)
\(c=lim\left[n^3\left(\frac{sin2n}{n^2}-3\right)\right]=+\infty\left(0-3\right)=-\infty\)
a) lim \(\frac{\left(2n+1\right)^2\left(n-1\right)}{\sqrt[3]{n^3+7n-2}}\)
= lim \(\left(2n+1\right)^2.\frac{\left(1-\frac{1}{n}\right)}{\sqrt[3]{1+\frac{7}{n^2}-\frac{2}{n^3}}}\)
\(=+\infty\)
b) lim \(\left(2n-1\right)\sqrt{\frac{2n^2+5}{n^4+n^2+2}}\)
= lim \(\left(2-\frac{1}{n}\right)\sqrt{\frac{2+\frac{5}{n^2}}{1+\frac{1}{n^2}+\frac{2}{n^4}}}\)
=2.2 = 4
c ) = lim \(n.\frac{n^2}{\sqrt[3]{\left(n^3+n^2\right)^2+n\sqrt[3]{n^3+n^2}+n^2}}\)
= lim \(n.\frac{1}{\sqrt[3]{\left(1+\frac{1}{n}\right)^2+\sqrt[3]{1+\frac{1}{n}}+1}}\)
\(=+\infty\)
\(a=\lim\limits n\left(\sqrt[3]{\frac{1}{n}+1}+1\right)=+\infty.2=+\infty\)
\(b=\lim\limits\frac{n^2+2\sqrt{n}+3}{2n^2+n-\sqrt{n}}=\lim\limits\frac{1+\frac{2}{n\sqrt{n}}+\frac{3}{n^2}}{2+\frac{1}{n}-\frac{1}{n\sqrt{n}}}=\frac{1}{2}\)
\(c=\lim\limits\frac{2n\sqrt{n}+3}{n^2+n+1}=\frac{\frac{2}{\sqrt{n}}+\frac{3}{n^2}}{1+\frac{1}{n}+\frac{1}{n^2}}=\frac{0}{1}=0\)
\(d=\lim\limits\frac{2n^2+6n\sqrt{n}}{n^2+3n+2}=\lim\limits\frac{2+\frac{6}{\sqrt{n}}}{1+\frac{3}{n}+\frac{2}{n^2}}=\frac{2}{1}=2\)
a/ \(=lim\frac{1}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\infty}=0\)
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: \(1^2+2^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)
\(\Rightarrow lim\frac{n\left(n+1\right)\left(2n+1\right)}{6n\left(n+1\right)\left(n+2\right)}=lim\frac{2n+1}{6n+12}=lim\frac{2+\frac{1}{n}}{6+\frac{12}{n}}=\frac{2}{6}=\frac{1}{3}\)
\(\lim\limits\dfrac{\sqrt{2\cdot4^n+1}-2^n}{\sqrt{2\cdot4^n+1}+2^n}\)
\(=\lim\limits\dfrac{2^n\cdot\sqrt{2+\dfrac{1}{4^n}}-2^n}{2^n\cdot\sqrt{2+\dfrac{1}{4^n}}+2^n}\)
\(=\lim\limits\dfrac{\sqrt{2+\dfrac{1}{4^n}}-1}{\sqrt{2+\dfrac{1}{4^n}}+1}=\dfrac{\sqrt{2}-1}{\sqrt{2}+1}\)
\(=\dfrac{\left(\sqrt{2}-1\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}=\dfrac{3-2\sqrt{2}}{2-1}=3-2\sqrt{2}\)
=>a=3; b=-2
\(a^3+b^3=3^3+\left(-2\right)^3=27-8=19\)