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NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 2 2021
\(a=\lim\dfrac{5n\left(n+\sqrt{n^2-n-1}\right)}{n+1}=\lim\dfrac{5\left(n+\sqrt{n^2-n-1}\right)}{1+\dfrac{1}{n}}=\dfrac{+\infty}{1}=+\infty\)
\(b=\lim\dfrac{\sqrt{\dfrac{1}{n}+\sqrt{\dfrac{1}{n^3}+\dfrac{1}{n^4}}}}{1-\dfrac{1}{\sqrt{n}}}=\dfrac{0}{1}=0\)
\(c=\lim\dfrac{\sqrt{2n^2-1+\dfrac{7}{n^2}}}{3+\dfrac{5}{n}}=\dfrac{+\infty}{3}=+\infty\)
\(d=\lim\dfrac{\sqrt{3+\dfrac{2}{n}}-1}{3-\dfrac{2}{n}}=\dfrac{\sqrt{3}-1}{3}\)
Y
15 tháng 10 2023
\(1,\lim\limits_{n\rightarrow\infty}\dfrac{-n^2+2n+1}{\sqrt{3n^4+2}}\left(1\right)\)
\(\dfrac{-n^2+2n+1}{\sqrt{3n^4+2}}=\dfrac{-\dfrac{n^2}{n^4}+\dfrac{2n}{n^4}+\dfrac{1}{n^4}}{\sqrt{\dfrac{3n^4}{n^4}+\dfrac{2}{n^4}}}=\dfrac{-\dfrac{1}{n^2}+\dfrac{2}{n^3}+\dfrac{1}{n^4}}{\sqrt{3+\dfrac{2}{n^4}}}\)
\(\Rightarrow\left(1\right)=\dfrac{-lim\dfrac{1}{n^2}+2lim\dfrac{1}{n^3}+lim\dfrac{1}{n^4}}{\sqrt{lim\left(3+\dfrac{2}{n^4}\right)}}\)
\(=\dfrac{0}{\sqrt{lim\left(3+\dfrac{2}{n^4}\right)}}=0\)
\(2,\lim\limits_{n\rightarrow\infty}\left(\dfrac{4n-\sqrt{16n^2+1}}{n+1}\right)\left(2\right)\)
\(\dfrac{4n-\sqrt{16n^2+1}}{n+1}=\dfrac{\dfrac{4n}{n^2}-\sqrt{\dfrac{16n^2}{n^2}+\dfrac{1}{n^2}}}{\dfrac{n}{n^2}+\dfrac{1}{n^2}}=\dfrac{\dfrac{4}{n}-\sqrt{16+\dfrac{1}{n^2}}}{\dfrac{1}{n}+\dfrac{1}{n^2}}\)
\(\Rightarrow\left(2\right)=\dfrac{lim\left(\dfrac{4}{n}-\sqrt{16+\dfrac{1}{n^2}}\right)}{lim\left(\dfrac{1}{n}+\dfrac{1}{n^2}\right)}=\dfrac{lim\left(\dfrac{4}{n}-\sqrt{16+\dfrac{1}{n^2}}\right)}{0}\)
Vậy giới hạn \(\left(2\right)\) không xác định.
\(3,\lim\limits_{n\rightarrow\infty}\left(\dfrac{\sqrt{9n^2+n+1}-3n}{2n}\right)\left(3\right)\)
\(\dfrac{\sqrt{9n^2+n+1}-3n}{2n}=\dfrac{\sqrt{9+\dfrac{1}{n}+\dfrac{1}{n^2}}-\dfrac{3}{n}}{\dfrac{2}{n}}\)
\(\Rightarrow\left(3\right)=\dfrac{lim\left(\sqrt{9+\dfrac{1}{n}+\dfrac{1}{n^2}}-\dfrac{3}{n}\right)}{2lim\dfrac{1}{n}}=\dfrac{lim\left(\sqrt{9+\dfrac{1}{n}+\dfrac{1}{n^2}}-\dfrac{3}{n}\right)}{0}\)
Vậy \(lim\left(3\right)\) không xác định.
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 2 2022
\(\lim\dfrac{n\sqrt{1+2+...+2n}}{3n^2+n-2}=\lim\dfrac{n\sqrt{\dfrac{2n\left(2n+1\right)}{2}}}{3n^2+n-2}=\lim\dfrac{\sqrt{2+\dfrac{1}{n}}}{3+\dfrac{1}{n}-\dfrac{2}{n^2}}=\dfrac{\sqrt{2}}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 2 2021
\(a=\lim n\left(\sqrt[3]{-1+\dfrac{2}{n}-\dfrac{5}{n^3}}\right)=+\infty.\left(-1\right)=-\infty\)
\(b=\lim\left(\sqrt{n+1}+\sqrt{n}\right)=+\infty\)
\(c=\lim n\left(\dfrac{1}{n^2+n}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(d=\lim\left(\dfrac{2n^2-1-2n\left(n+1\right)}{n+1}\right)=\lim\left(\dfrac{-1-2n}{n+1}\right)=-2\)
\(e=\lim\dfrac{2n^2+n-3+\dfrac{1}{n}}{\dfrac{2}{n}-3}=\dfrac{+\infty}{-3}=-\infty\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 2 2021
\(a=\lim\dfrac{-2n^2}{\sqrt{n^2+2}+\sqrt{n^2+4}}=\lim\dfrac{-2n}{\sqrt{1+\dfrac{2}{n^2}}+\sqrt{1+\dfrac{4}{n^2}}}=\dfrac{-\infty}{2}=-\infty\)
\(b=\lim\dfrac{3-5n^2+10n}{n-2}=\lim\dfrac{-5n+10+\dfrac{3}{n}}{1-\dfrac{2}{n}}=\dfrac{-\infty}{1}=-\infty\)
\(c=\lim\left(\dfrac{1-\dfrac{1}{n}}{\dfrac{\sqrt{3}}{n}-1}-4.2^n\right)=-1-\infty=-\infty\)
\(d=\lim\dfrac{n^3-4n-\left(3n^2+4\right)\left(n-2\right)}{n^2-2n}=\lim\dfrac{-2n^3+6n^2-8n+8}{n^2-2n}\)
\(\lim\dfrac{-2n+6-\dfrac{8}{n}+\dfrac{8}{n^2}}{1-\dfrac{2}{n}}=\dfrac{-\infty}{1}=-\infty\)
\(e=\lim\dfrac{\sqrt{1+\dfrac{1}{n}}-\sqrt{5}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{5}}=\dfrac{1-\sqrt{5}}{1+\sqrt{5}}\)
S
1
15 tháng 11 2023
\(\lim\limits\left(\sqrt{n^2+2n}-\sqrt{n^2-2n}\right)\)
\(=\lim\limits\dfrac{n^2+2n-n^2+2n}{\sqrt{n^2+2n}+\sqrt{n^2-2n}}\)
\(=\lim\limits\dfrac{4n}{\sqrt{n^2+2n}+\sqrt{n^2-2n}}\)
\(=\lim\limits\dfrac{4}{\sqrt{1+\dfrac{2}{n}}+\sqrt{1-\dfrac{2}{n}}}\)
\(=\dfrac{4}{1+1}=\dfrac{4}{2}=2\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 1 2021
\(=\lim\left(\dfrac{1+2+...+n-1}{n^2}\right)=\lim\dfrac{n\left(n-1\right)}{2n^2}=\dfrac{1}{2}\)
AM
11 tháng 2 2022
a. ĐKXĐ: \(n\ge0\)
\(lim_{n\rightarrow0}\dfrac{\sqrt{2n+1}}{\sqrt{8n}+1}=\dfrac{\sqrt{2.0+1}}{\sqrt{8.0}+1}=1\)
\(lim_{n\rightarrow+\infty}\dfrac{\sqrt{2n+1}}{\sqrt{8n}+1}=lim_{n\rightarrow+\infty}\dfrac{\sqrt{2+\dfrac{1}{n}}}{\sqrt{8}+\dfrac{1}{\sqrt{n}}}=\dfrac{1}{2}\)
b. ĐKXĐ: \(\left\{{}\begin{matrix}n\ne0\\n\le\dfrac{-1-\sqrt{21}}{2}\\n\ge\dfrac{-1+\sqrt{21}}{2}\end{matrix}\right.\)
\(lim_{n\rightarrow+\infty}\dfrac{3n+\sqrt{n^2+n-5}}{-2n}=\)\(lim_{n\rightarrow+\infty}\dfrac{3+\sqrt{1+\dfrac{1}{n}-\dfrac{5}{n^2}}}{-2}=-2\)
\(lim_{n\rightarrow-\infty}\dfrac{3n+\sqrt{n^2+n-5}}{-2n}=\)\(lim_{n\rightarrow-\infty}\dfrac{3+\sqrt{1+\dfrac{1}{n}-\dfrac{5}{n^2}}}{-2}=-1\)
HP
18 tháng 2 2022
a, \(lim\dfrac{\sqrt{2n+1}}{\sqrt{8n}+1}=lim\dfrac{\sqrt{n}.\sqrt{2+\dfrac{1}{n}}}{\sqrt{n}\left(\sqrt{8}+\dfrac{1}{n}\right)}=\dfrac{\sqrt{2}}{\sqrt{8}}=\dfrac{1}{2}\)
`lim[n\sqrt{n}+1]/[n^2+2]`
`=lim[n^2\sqrt{1/n}+1]/[n^2+2]`
`=lim[n^2(\sqrt{1/n}+1/[n^2])]/[n^2(1+2/[n^2])]`
`=lim[\sqrt{1/n}+1/[n^2]]/[1+2/[n^2]]`
`=0/1=0`