tìm \(lim\left(8-2n-\sqrt[3]{6n-8n^3}\right)\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
MH
11 tháng 2 2022
\(a,lim\dfrac{^3\sqrt{8n^3+2n}}{-n+3}\)
\(=lim\dfrac{^3\sqrt{8+\dfrac{2}{n^2}}}{-1+\dfrac{3}{n}}=\dfrac{^3\sqrt{8}}{-1}=\dfrac{2}{-1}=-2\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 2 2022
\(\lim\dfrac{\left(2n\sqrt{n}+1\right)\left(\sqrt{n}+3\right)}{\left(n-1\right)\left(3-2n\right)}=\lim\dfrac{\left(2+\dfrac{1}{n\sqrt{n}}\right)\left(1+\dfrac{3}{\sqrt{n}}\right)}{\left(1-\dfrac{1}{n}\right)\left(\dfrac{3}{n}-2\right)}=\dfrac{2.1}{1.\left(-2\right)}=-1\)
ST
1
4 tháng 11 2023
a: \(\lim\limits\dfrac{5n+1}{2n}=\lim\limits\dfrac{\dfrac{5n}{n}+\dfrac{1}{n}}{\dfrac{2n}{n}}=\lim\limits\dfrac{5+\dfrac{1}{n}}{2}=\dfrac{5+0}{2}=\dfrac{5}{2}\)
b: \(\lim\limits\dfrac{6n^2+8n+1}{5n^2+3}\)
\(=\lim\limits\dfrac{\dfrac{6n^2}{n^2}+\dfrac{8n}{n^2}+\dfrac{1}{n^2}}{\dfrac{5n^2}{n^2}+\dfrac{3}{n^2}}\)
\(=\lim\limits\dfrac{6+\dfrac{8}{n}+\dfrac{1}{n^2}}{5+\dfrac{3}{n^2}}\)
\(=\dfrac{6+0+0}{5+0}=\dfrac{6}{5}\)
c: \(\lim\limits\dfrac{3^n+2^n}{4\cdot3^n}\)
\(=\lim\limits\dfrac{\dfrac{3^n}{3^n}+\left(\dfrac{2}{3}\right)^n}{4\cdot\left(\dfrac{3^n}{3^n}\right)}\)
\(=\lim\limits\dfrac{1+\left(\dfrac{2}{3}\right)^n}{4}=\dfrac{1+0}{4}=\dfrac{1}{4}\)
d: \(\lim\limits\dfrac{\sqrt{n^2+5n+3}}{6n+2}\)
\(=\lim\limits\dfrac{\sqrt{\dfrac{n^2}{n^2}+\dfrac{5n}{n^2}+\dfrac{3}{n^2}}}{\dfrac{6n}{n}+\dfrac{2}{n}}\)
\(=\lim\limits\dfrac{\sqrt{1+\dfrac{5}{n}+\dfrac{3}{n^2}}}{6+\dfrac{2}{n}}\)
\(=\dfrac{\sqrt{1+0+0}}{6}=\dfrac{1}{6}\)
AD
4 tháng 11 2023
\(a,lim\dfrac{5n+1}{2n}=lim\dfrac{\dfrac{5n}{n}+\dfrac{1}{n}}{\dfrac{2n}{n}}=lim\dfrac{5+\dfrac{1}{n}}{2}=\dfrac{5}{2}\\ b,lim\dfrac{6n^2+8n+1}{5n^2+3}=lim\dfrac{\dfrac{6n^2}{n^2}+\dfrac{8n}{n^2}+\dfrac{1}{n^2}}{\dfrac{5n^2}{n^2}+\dfrac{3}{n^2}}=lim\dfrac{6+\dfrac{8}{n}+\dfrac{1}{n^2}}{5+\dfrac{3}{n^2}}=\dfrac{6}{5}\)
\(c,lim\dfrac{3^n+2^n}{4.3^n}=\dfrac{\dfrac{3^n}{3^n}+\dfrac{2^n}{3^n}}{\dfrac{4.3^n}{3^n}}=\dfrac{1+\left(\dfrac{2}{3}\right)^n}{4}=\dfrac{1}{4}\)
\(d,lim\dfrac{\sqrt{n^2+5n+3}}{6n+2}=lim\dfrac{\sqrt{\dfrac{n^2+5n+3}{n^2}}}{\dfrac{6n}{n}+\dfrac{2}{n}}=lim\dfrac{\sqrt{1+\dfrac{5}{n}+\dfrac{3}{n^2}}}{6+\dfrac{2}{n}}=\dfrac{1}{6}\)
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 \(\lim \left( {2 + \frac{1}{n}} \right) = \lim 2 + \lim \frac{1}{n} = 2 + 0 = 2 > 0;\lim {3^n} = + \infty \Rightarrow \lim \frac{{2 + \frac{1}{n}}}{{{3^n}}} = 0\)
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.
\(\lim\left(\sqrt[3]{8n^3+3n^2-2}+\sqrt[3]{5n^2-8n^3}\right)\)
\(=\lim\left(\sqrt[3]{8n^3+3n^2-2}-\sqrt[3]{8n^3-5n^2}\right)\)
\(=\lim\dfrac{8n^3+3n^2-2-\left(8n^3-5n^2\right)}{\sqrt[3]{\left(8n^3+3n^2-2\right)^2}+\sqrt[3]{\left(8n^3+3n^2-2\right)\left(8n^3-5n^2\right)}+\sqrt[3]{8n^3-5n^2}}\)
\(=\lim\dfrac{8n^2-2}{\sqrt[3]{\left(8n^3+3n^2-2\right)^2}+\sqrt[3]{\left(8n^3+3n^2-2\right)\left(8n^3-5n^2\right)}+\sqrt[3]{8n^3-5n^2}}\)
\(=lim\dfrac{8-\dfrac{2}{n^2}}{\sqrt[3]{\left(8+\dfrac{3}{n}-\dfrac{2}{n^3}\right)^2}+\sqrt[3]{\left(8+\dfrac{3}{n}-\dfrac{2}{n^3}\right)\left(8-\dfrac{5}{n}\right)}+\sqrt[3]{\left(8-\dfrac{5}{n}\right)^2}}\)
\(=\dfrac{8}{\sqrt[3]{8^2}+\sqrt[3]{8.8}+\sqrt[3]{8^2}}=...\)