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1) = lim n. \(\frac{n^3-3n^2-27n^3}{\sqrt[3]{\left(n^3-3n^2\right)^2}+3n\sqrt[3]{n^3-3n^2}+9n^2}\)
= lim \(\frac{n\left(-26n^3-3n^2\right)}{\sqrt[3]{\left(n^3-3n^2\right)^2}+3n\sqrt[3]{n^3-3n^2}+9n^2}\)
= lim \(\frac{n^2\left(-26-\frac{3}{n}\right)}{\sqrt[3]{\left(1-\frac{3}{n}\right)^2}+3\sqrt[3]{1-\frac{3}{n}}+9}\)
= lim \(\frac{n^2\left(-26\right)}{13}=-\infty\)
2) = lim ( \(\sqrt{4n^2+n}-2n+\sqrt[3]{2n^2-8n^3}+2n\))
= lim ( \(\frac{n}{\sqrt{4n^2+n}+2n}+\frac{2n^2}{\sqrt[3]{\left(2n^2-8n^3\right)^2}-2n\sqrt[3]{2n^2-8n^3}+4n^2}\))
= \(\frac{1}{2+2}+\frac{2}{4+4+4}=\frac{5}{12}\)
\(=lim\left(\sqrt{4n^2+2n}-2n+n-\sqrt[3]{n^3+2n-1}+4\right)\)
\(=lim\left(\frac{2n}{\sqrt{4n^2+2n}+2n}+\frac{-2n+1}{\sqrt[3]{\left(n^3+2n-1\right)^2}+n\sqrt[3]{n^3+2n-1}+n^2}+4\right)\)
\(=lim\left(\frac{2}{\sqrt{4+\frac{2}{n}}+2}+\frac{-2+\frac{1}{n}}{\sqrt[3]{\left(n^{\frac{3}{2}}+2.n^{-\frac{1}{2}}-n^{-\frac{3}{2}}\right)}+\sqrt[3]{n^3+2n-1}+n}+4\right)\)
\(=\frac{2}{2+2}+0+4=\frac{9}{2}\)
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\(\frac{2+\frac{3}{n}}{\sqrt{9+\frac{3}{n^2}}-\sqrt[3]{\frac{2}{n}-8}}=\frac{2}{3+2}=\frac{2}{5}\)
\(lim\dfrac{5n\sqrt{2n^2-n}}{1+5n-3n^2}=lim\dfrac{5\sqrt{2-\dfrac{1}{n}}}{\dfrac{1}{n^2}+\dfrac{5}{n}-3}=\dfrac{5\sqrt{2-0}}{0+0-3}=\dfrac{-5\sqrt{2}}{3}\)
\(lim\dfrac{\sqrt{4n^2+n}-7n}{3n^2-1}=lim\dfrac{\sqrt{\dfrac{4}{n^2}+\dfrac{1}{n^3}}-\dfrac{7}{n}}{3-\dfrac{1}{n^2}}=\dfrac{\sqrt{0+0}-0}{3-0}=\dfrac{0}{3}=0\)
a)lim \(\frac{\sqrt{n^2-4n}-\sqrt{4n+1}}{\sqrt{3n^2+1}+n}\)
=lim \(\frac{\sqrt{1-\frac{4}{n}}-\sqrt{\frac{4}{n}+\frac{1}{n^2}}}{\sqrt{3+\frac{1}{n^2}}+1}=\frac{1}{\sqrt{3}+1}\)
b)lim \(\frac{\sqrt[3]{8n^3+n^2}-n}{2n-3}\)
= lim \(\frac{\sqrt[3]{8+\frac{1}{n^3}}-1}{2-\frac{3}{n}}=\frac{2-1}{2}=\frac{1}{2}\)
\(\lim\limits\left(\sqrt{2n^2+3}-\sqrt{n^2+1}\right)=\lim\limits\frac{n^2-2}{\left(\sqrt{2n^2+3}+\sqrt{n^2+1}\right)}=\lim\limits\frac{n-\frac{2}{n}}{\sqrt{2+\frac{3}{n^2}}+\sqrt{1+\frac{1}{n^2}}}=+\infty\)
\(\lim\limits\frac{1}{\sqrt{n+1}-\sqrt{n}}=\lim\limits\left(\sqrt{n+1}+\sqrt{n}\right)=+\infty\)
\(=\lim\dfrac{\sqrt{4-\dfrac{1}{n}}+\sqrt[3]{8+\dfrac{1}{n}}}{2+\dfrac{3}{n}}=\dfrac{2+2}{2}=2\)