Bạn chưa đăng nhập. Vui lòng đăng nhập để hỏi bài

Những câu hỏi liên quan
Chuột yêu Gạo
Xem chi tiết
Minh Hiếu
12 tháng 2 2022 lúc 20:10

\(a,lim\left(8n-3n^9+1\right)\)

\(=limn^9\left(\dfrac{8}{n^8}-3+\dfrac{1}{n^9}\right)\)

\(=n^9\left(0-3+0\right)=n^9.\left(-3\right)=\)-∞

 

Nguyễn Việt Lâm
12 tháng 2 2022 lúc 20:34

\(\lim\left(6n^4-n+1\right)=\lim n^4\left(6-\dfrac{1}{n^3}+\dfrac{1}{n^4}\right)=+\infty.6=+\infty\)

\(\lim\left(2-3n+7n^2\right)=\lim n^2\left(\dfrac{2}{n^2}-\dfrac{3}{n}+7\right)=+\infty.7=+\infty\)

Nguyễn Trọng Hùng
Xem chi tiết
Hồng Phúc
22 tháng 3 2022 lúc 16:29

undefined

Julian Edward
Xem chi tiết
Nguyễn Việt Lâm
6 tháng 2 2021 lúc 20:37

\(a=\lim\dfrac{1}{\sqrt{4n+1}+2\sqrt{n}}=\dfrac{1}{\infty}=0\)

\(b=\lim n\left(\sqrt{1+\dfrac{2}{n}}-\sqrt{1-\dfrac{2}{n}}-1\right)=+\infty.\left(-1\right)=-\infty\)

\(c=\lim4^n\left(\sqrt{\left(\dfrac{9}{16}\right)^n-\left(\dfrac{3}{16}\right)^n}-1\right)=+\infty.\left(-1\right)=-\infty\)

\(d=\lim n^3\left(3+\dfrac{2}{n}+\dfrac{1}{n^2}\right)=+\infty.3=+\infty\)

Đừng gọi tôi là Jung Hae...
Xem chi tiết
Minh Hiếu
11 tháng 2 2022 lúc 5:34

\(a,lim\dfrac{^3\sqrt{8n^3+2n}}{-n+3}\)

\(=lim\dfrac{^3\sqrt{8+\dfrac{2}{n^2}}}{-1+\dfrac{3}{n}}=\dfrac{^3\sqrt{8}}{-1}=\dfrac{2}{-1}=-2\)

Nguyễn Việt Lâm
12 tháng 2 2022 lúc 21:02

\(\lim\dfrac{\left(2n\sqrt{n}+1\right)\left(\sqrt{n}+3\right)}{\left(n-1\right)\left(3-2n\right)}=\lim\dfrac{\left(2+\dfrac{1}{n\sqrt{n}}\right)\left(1+\dfrac{3}{\sqrt{n}}\right)}{\left(1-\dfrac{1}{n}\right)\left(\dfrac{3}{n}-2\right)}=\dfrac{2.1}{1.\left(-2\right)}=-1\)

Măm Măm
Xem chi tiết
Nguyễn Việt Lâm
13 tháng 2 2022 lúc 21:59

\(\lim\left(\sqrt{4n^2+5n}-2n\right)=\lim\dfrac{5n}{\sqrt{4n^2+5n}+2n}=\lim\dfrac{5}{\sqrt{4+\dfrac{5}{n}}+2}=\dfrac{5}{\sqrt{4+0}+2}=\dfrac{5}{4}\)

\(\lim\left(\sqrt{2n+1}-\sqrt{n}\right)=\lim\sqrt{n}\left(\sqrt{2+\dfrac{1}{n}}-1\right)=+\infty.\left(\sqrt{2}-1\right)=+\infty\) (do \(\sqrt{2}-1>0\))

Minh Hiếu
13 tháng 2 2022 lúc 22:00

\(a,lim\left(\sqrt{4n^2+5n}-2n\right)\)

\(=limn\left(\sqrt{4+\dfrac{5}{n}}-2\right)=n.0=0\)

\(b,lim\left(\sqrt{2n+1}-\sqrt{n}\right)\)

\(=lim\sqrt{n}\left(\sqrt{2+\dfrac{1}{n}}-1\right)=\sqrt{n}\left(\sqrt{2}-1\right)=+\infty\)

đoàn ngọc hân
Xem chi tiết
Nguyễn Việt Lâm
17 tháng 1 2021 lúc 13:22

\(a=\lim\left(\dfrac{2n^3\left(5n+1\right)+\left(2n^2+3\right)\left(1-5n^2\right)}{\left(2n^2+3\right)\left(5n+1\right)}\right)\)

\(=\lim\left(\dfrac{2n^3-13n^2+3}{\left(2n^2+3\right)\left(5n+1\right)}\right)=\lim\dfrac{2-\dfrac{13}{n}+\dfrac{3}{n^3}}{\left(2+\dfrac{3}{n^2}\right)\left(5+\dfrac{1}{n}\right)}=\dfrac{2}{2.5}=\dfrac{1}{5}\)

\(b=\lim\left(\dfrac{n-2}{\sqrt{n^2+n}+\sqrt{n^2+2}}\right)=\lim\dfrac{1-\dfrac{2}{n}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1+\dfrac{2}{n}}}=\dfrac{1}{2}\)

\(c=\lim\dfrac{\sqrt{1+\dfrac{3}{n^3}-\dfrac{2}{n^4}}}{2-\dfrac{2}{n}+\dfrac{3}{n^2}}=\dfrac{1}{2}\)

\(d=\lim\dfrac{\sqrt{1-\dfrac{4}{n}}-\sqrt{4+\dfrac{1}{n^2}}}{\sqrt{3+\dfrac{1}{n^2}}-1}=\dfrac{1-2}{\sqrt{3}-1}=-\dfrac{1+\sqrt{3}}{2}\)

Dương thị bầu
15 tháng 3 2022 lúc 20:57

Lim 3.4n-2.13n/5n+6.13n

Măm Măm
Xem chi tiết
Nguyễn Việt Lâm
13 tháng 2 2022 lúc 22:06

\(\lim\left(3n-\sqrt{9n^2+1}\right)=\lim\dfrac{-1}{3n+\sqrt{9n^2+1}}=\lim\dfrac{-\dfrac{1}{n}}{3+\sqrt{9+\dfrac{1}{n^2}}}=\dfrac{0}{3+3}=0\)

\(\lim\left(\sqrt[3]{n^3-2n^2}-n\right)=\lim\dfrac{-2n^2}{\sqrt[3]{\left(n^3-2n^2\right)^2}+n\sqrt[3]{n^3-2n^2}+n^2}\)

\(=\lim\dfrac{-2}{\sqrt[3]{\left(1-\dfrac{2}{n}\right)^2}+\sqrt[3]{1-\dfrac{2}{n}}+1}=\dfrac{-2}{1+1+1}=-\dfrac{2}{3}\)

ánh tuyết nguyễn
Xem chi tiết
2611
8 tháng 1 2023 lúc 14:37

`a)lim[2n^2+5]/[-3n^2-3]`

`=lim[2+5/[n^2]]/[-3-3/[n^2]]`

`=2/[-3]=-2/3`

`b)lim(-5n^3-2n^2+5n-6)`

`=lim n^3(-5-2/n+5/[n^2]-6/[n^3])`

Vì `{:(lim n^3=+oo),(lim (-5-2/n+5/[n^2]-6/[n^3])=-5):}}=>lim n^3(-5-2/n+5/[n^2]-6/[n^3])=-oo`

你混過 vulnerable 他 難...
Xem chi tiết
Nguyễn Việt Lâm
4 tháng 12 2021 lúc 15:49

\(\lim\limits\left(2-3n\right)^4\left(n+1\right)^3=\lim n^7\left(3-\dfrac{2}{n}\right)^4\left(1+\dfrac{1}{n}\right)^3=+\infty\)

\(\lim\left(\sqrt[3]{n+4}-\sqrt[3]{n+1}\right)=\lim\dfrac{3}{\sqrt[3]{\left(n+4\right)^2}+\sqrt[3]{\left(n+4\right)\left(n+1\right)}+\sqrt[3]{\left(n+1\right)^2}}=0\)

\(\lim\left(\sqrt[3]{8n^3+3n^2+4}-2n+6\right)=\lim\dfrac{8n^3+3n^2+4-\left(2n-6\right)^3}{\sqrt[3]{\left(8n^3+3n^2+4\right)^2}+\left(2n-6\right)\sqrt[3]{8n^3+3n^2+4}+\left(2n-6\right)^2}\)

\(=\lim\dfrac{75n^2-216n+220}{\sqrt[3]{\left(8n^3+3n^2+4\right)^2}+\left(2n-6\right)\sqrt[3]{8n^3+3n^2+4}+\left(2n-6\right)^2}\)

\(=\lim\dfrac{75-\dfrac{216}{n}+\dfrac{220}{n^2}}{\sqrt[3]{\left(8+\dfrac{3}{n}+\dfrac{4}{n^3}\right)^2}+\left(2-\dfrac{6}{n}\right)\sqrt[3]{8+\dfrac{3}{n}+\dfrac{4}{n^3}}+\left(2-\dfrac{6}{n}\right)^2}\)

\(=\dfrac{75}{\sqrt[3]{8^2}+2.\sqrt[3]{8}+2^2}=...\)

Nguyễn Việt Lâm
4 tháng 12 2021 lúc 15:52

d.

\(\lim\left(\sqrt[3]{8n^3+3n^2-2}+\sqrt[3]{5n^2-8n^3}\right)\)

\(=\lim\left(\sqrt[3]{8n^3+3n^2-2}-\sqrt[3]{8n^3-5n^2}\right)\)

\(=\lim\dfrac{8n^3+3n^2-2-\left(8n^3-5n^2\right)}{\sqrt[3]{\left(8n^3+3n^2-2\right)^2}+\sqrt[3]{\left(8n^3+3n^2-2\right)\left(8n^3-5n^2\right)}+\sqrt[3]{8n^3-5n^2}}\)

\(=\lim\dfrac{8n^2-2}{\sqrt[3]{\left(8n^3+3n^2-2\right)^2}+\sqrt[3]{\left(8n^3+3n^2-2\right)\left(8n^3-5n^2\right)}+\sqrt[3]{8n^3-5n^2}}\)

\(=lim\dfrac{8-\dfrac{2}{n^2}}{\sqrt[3]{\left(8+\dfrac{3}{n}-\dfrac{2}{n^3}\right)^2}+\sqrt[3]{\left(8+\dfrac{3}{n}-\dfrac{2}{n^3}\right)\left(8-\dfrac{5}{n}\right)}+\sqrt[3]{\left(8-\dfrac{5}{n}\right)^2}}\)

\(=\dfrac{8}{\sqrt[3]{8^2}+\sqrt[3]{8.8}+\sqrt[3]{8^2}}=...\)