KT
1
K
Khách
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KT
1
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 3 2019
\(lim\left(2n-\sqrt{8n^3+8n^2+2}\right)=lim\left(2n-2n^{\frac{3}{2}}\sqrt{2+\frac{2}{n}+\frac{1}{2n^2}}\right)\)
\(=lim\left(n\left(1-2\sqrt{n}\sqrt{2+\frac{2}{n}+\frac{1}{2n^2}}\right)\right)=\infty\times\left(-\infty\right)=-\infty\)
24 tháng 4 2020
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 ( \(\frac{n}{\sqrt{4n^2+n}+2n}+\frac{2n^2}{\sqrt[3]{\left(2n^2-8n^3\right)^2}-2n\sqrt[3]{2n^2-8n^3}+4n^2}\))
= \(\frac{1}{2+2}+\frac{2}{4+4+4}=\frac{5}{12}\)
24 tháng 4 2020
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\(\frac{2+\frac{3}{n}}{\sqrt{9+\frac{3}{n^2}}-\sqrt[3]{\frac{2}{n}-8}}=\frac{2}{3+2}=\frac{2}{5}\)
ST
0
1 tháng 4 2020
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 \(\frac{\sqrt[3]{8+\frac{1}{n^3}}-1}{2-\frac{3}{n}}=\frac{2-1}{2}=\frac{1}{2}\)
22 tháng 1 2020
Câu 1.
Vì \(\sqrt{2},\left(\sqrt{2}\right)^2,...,\left(\sqrt{2}\right)^n\) lập thành cấp số nhân có \(u_1=\sqrt{2}=q\) nên
\({u_n} = \sqrt 2 .\dfrac{{1 - {{\left( {\sqrt 2 } \right)}^n}}}{{1 - \sqrt 2 }} = \left( {2 - \sqrt 2 } \right)\left[ {{{\left( {\sqrt 2 } \right)}^n} - 1} \right] \to \lim {u_n} = + \infty \) vì \(\left\{{}\begin{matrix}a=2-\sqrt{2}>0\\q=\sqrt{2}>1\end{matrix}\right.\)
22 tháng 1 2020
Câu 3.
Ta có biến đổi:
\(\lim \left( {\dfrac{{{n^2} - n}}{{1 - 2{n^2}}} + \dfrac{{2\sin {n^2}}}{{\sqrt n }}} \right) = \lim \dfrac{{{n^2} - n}}{{1 - 2{n^2}}} = \dfrac{1}{2}\)
\(lim\left(2n-\sqrt[3]{8n^3+8n^2+2}\right)\)
\(=lim\frac{\left(2n-\sqrt[3]{8n^3+8n^2+2}\right)\left(4n^2+2n\sqrt[3]{8n^3+8n^2+2}+\sqrt[3]{\left(8n^3+8n^2+2\right)^2}\right)}{4n^2+2n\sqrt[3]{8n^3+8n^2+2}+\sqrt[3]{\left(8n^3+8n^2+2\right)^2}}\)
\(=lim\frac{8n^3-\left(8n^3+8n^2+2\right)}{4n^2+2n\sqrt[3]{8n^3+8n^2+2}+\sqrt[3]{\left(8n^3+8n^2+2\right)^2}}\)
\(=lim\frac{-8n^2-2}{4n^2+2n\sqrt[3]{8n^3+8n^2+2}+\sqrt[3]{\left(8n^3+8n^2+2\right)^2}}\)
\(=lim\frac{-8-\frac{2}{n^2}}{4+2\sqrt[3]{8+\frac{8}{n}+\frac{2}{n^3}}+\sqrt[3]{\left(8+\frac{8}{n}+\frac{2}{n^3}\right)^2}}\)
\(=\frac{-8+0}{4+2\sqrt[3]{8+8+0}+\sqrt[3]{\left(8+0+0\right)^2}}=\frac{-2}{3}\)