K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Dưới đây là một vài câu hỏi có thể liên quan tới câu hỏi mà bạn gửi lên. Có thể trong đó có câu trả lời mà bạn cần!
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 3 2019
\(lim\left(2n-\sqrt{8n^3+8n^2+2}\right)=lim\left(2n-2n^{\frac{3}{2}}\sqrt{2+\frac{2}{n}+\frac{1}{2n^2}}\right)\)
\(=lim\left(n\left(1-2\sqrt{n}\sqrt{2+\frac{2}{n}+\frac{1}{2n^2}}\right)\right)=\infty\times\left(-\infty\right)=-\infty\)
QA
1
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 1 2022
\(=\lim\left(\sqrt[]{4n^2+2n+1}-2n+2n-\sqrt[3]{8n^3-3n^2+1}\right)\)
\(=\lim\left(\dfrac{2n+1}{\sqrt[]{4n^2+2n+1}+2n}+\dfrac{3n^2-1}{4n^2+2n\sqrt[3]{8n^3-3n^2+1}+\sqrt[3]{\left(8n^3-3n^2+1\right)^2}}\right)\)
\(=\lim\left(\dfrac{2+\dfrac{1}{n}}{\sqrt[]{4+\dfrac{2}{n}+\dfrac{1}{n^2}}+2}+\dfrac{3-\dfrac{1}{n^2}}{4+2\sqrt[3]{8-\dfrac{3}{n}+\dfrac{1}{n^3}}+\sqrt[3]{\left(8-\dfrac{3}{n}+\dfrac{1}{n^3}\right)^2}}\right)\)
\(=\dfrac{2}{\sqrt[]{4}+2}+\dfrac{3}{4+2\sqrt[3]{8}+\sqrt[3]{8^2}}=...\)
BD
1
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 1 2021
\(=\lim\dfrac{\sqrt{4-\dfrac{1}{n}}+\sqrt[3]{8+\dfrac{1}{n}}}{2+\dfrac{3}{n}}=\dfrac{2+2}{2}=2\)
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 12 2021
\(\lim\limits\left(2-3n\right)^4\left(n+1\right)^3=\lim n^7\left(3-\dfrac{2}{n}\right)^4\left(1+\dfrac{1}{n}\right)^3=+\infty\)
\(\lim\left(\sqrt[3]{n+4}-\sqrt[3]{n+1}\right)=\lim\dfrac{3}{\sqrt[3]{\left(n+4\right)^2}+\sqrt[3]{\left(n+4\right)\left(n+1\right)}+\sqrt[3]{\left(n+1\right)^2}}=0\)
\(\lim\left(\sqrt[3]{8n^3+3n^2+4}-2n+6\right)=\lim\dfrac{8n^3+3n^2+4-\left(2n-6\right)^3}{\sqrt[3]{\left(8n^3+3n^2+4\right)^2}+\left(2n-6\right)\sqrt[3]{8n^3+3n^2+4}+\left(2n-6\right)^2}\)
\(=\lim\dfrac{75n^2-216n+220}{\sqrt[3]{\left(8n^3+3n^2+4\right)^2}+\left(2n-6\right)\sqrt[3]{8n^3+3n^2+4}+\left(2n-6\right)^2}\)
\(=\lim\dfrac{75-\dfrac{216}{n}+\dfrac{220}{n^2}}{\sqrt[3]{\left(8+\dfrac{3}{n}+\dfrac{4}{n^3}\right)^2}+\left(2-\dfrac{6}{n}\right)\sqrt[3]{8+\dfrac{3}{n}+\dfrac{4}{n^3}}+\left(2-\dfrac{6}{n}\right)^2}\)
\(=\dfrac{75}{\sqrt[3]{8^2}+2.\sqrt[3]{8}+2^2}=...\)
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 12 2021
d.
\(\lim\left(\sqrt[3]{8n^3+3n^2-2}+\sqrt[3]{5n^2-8n^3}\right)\)
\(=\lim\left(\sqrt[3]{8n^3+3n^2-2}-\sqrt[3]{8n^3-5n^2}\right)\)
\(=\lim\dfrac{8n^3+3n^2-2-\left(8n^3-5n^2\right)}{\sqrt[3]{\left(8n^3+3n^2-2\right)^2}+\sqrt[3]{\left(8n^3+3n^2-2\right)\left(8n^3-5n^2\right)}+\sqrt[3]{8n^3-5n^2}}\)
\(=\lim\dfrac{8n^2-2}{\sqrt[3]{\left(8n^3+3n^2-2\right)^2}+\sqrt[3]{\left(8n^3+3n^2-2\right)\left(8n^3-5n^2\right)}+\sqrt[3]{8n^3-5n^2}}\)
\(=lim\dfrac{8-\dfrac{2}{n^2}}{\sqrt[3]{\left(8+\dfrac{3}{n}-\dfrac{2}{n^3}\right)^2}+\sqrt[3]{\left(8+\dfrac{3}{n}-\dfrac{2}{n^3}\right)\left(8-\dfrac{5}{n}\right)}+\sqrt[3]{\left(8-\dfrac{5}{n}\right)^2}}\)
\(=\dfrac{8}{\sqrt[3]{8^2}+\sqrt[3]{8.8}+\sqrt[3]{8^2}}=...\)
CC
1
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\(\frac{2+\frac{3}{n}}{\sqrt{9+\frac{3}{n^2}}-\sqrt[3]{\frac{2}{n}-8}}=\frac{2}{3+2}=\frac{2}{5}\)
HQ
Hà Quang Minh
Giáo viên
22 tháng 9 2023
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 \(\lim \left( {2 + \frac{1}{n}} \right) = \lim 2 + \lim \frac{1}{n} = 2 + 0 = 2 > 0;\lim {3^n} = + \infty \Rightarrow \lim \frac{{2 + \frac{1}{n}}}{{{3^n}}} = 0\)
\(lim\left(2n-\sqrt[3]{8n^3+8n^2+2}\right)\)
\(=lim\frac{\left(2n-\sqrt[3]{8n^3+8n^2+2}\right)\left(4n^2+2n\sqrt[3]{8n^3+8n^2+2}+\sqrt[3]{\left(8n^3+8n^2+2\right)^2}\right)}{4n^2+2n\sqrt[3]{8n^3+8n^2+2}+\sqrt[3]{\left(8n^3+8n^2+2\right)^2}}\)
\(=lim\frac{8n^3-\left(8n^3+8n^2+2\right)}{4n^2+2n\sqrt[3]{8n^3+8n^2+2}+\sqrt[3]{\left(8n^3+8n^2+2\right)^2}}\)
\(=lim\frac{-8n^2-2}{4n^2+2n\sqrt[3]{8n^3+8n^2+2}+\sqrt[3]{\left(8n^3+8n^2+2\right)^2}}\)
\(=lim\frac{-8-\frac{2}{n^2}}{4+2\sqrt[3]{8+\frac{8}{n}+\frac{2}{n^3}}+\sqrt[3]{\left(8+\frac{8}{n}+\frac{2}{n^3}\right)^2}}\)
\(=\frac{-8+0}{4+2\sqrt[3]{8+8+0}+\sqrt[3]{\left(8+0+0\right)^2}}=\frac{-2}{3}\)