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HT
1 tháng 2 2021
a/ \(=\lim\limits\dfrac{\sqrt{\dfrac{n}{n}+\dfrac{1}{n}}}{\dfrac{1}{\sqrt{n}}+\sqrt{\dfrac{n}{n}}}=1\)
b/ \(1+2+...+n=\dfrac{n\left(n+1\right)}{2}\)
\(\Rightarrow\lim\limits\dfrac{n\left(n+1\right)}{2n^2+4}=\lim\limits\dfrac{\dfrac{n^2}{n^2}+\dfrac{n}{n^2}}{\dfrac{2n^2}{n^2}+\dfrac{4}{n^2}}=\dfrac{1}{2}\)
c/ \(=\lim\limits\dfrac{n^2+n+1-n^2}{\sqrt{n^2+n+1}+n}=\lim\limits\dfrac{n+1}{\sqrt{n^2+n+1}+n}=\lim\limits\dfrac{\dfrac{n}{n}+\dfrac{1}{n}}{\sqrt{\dfrac{n^2}{n^2}+\dfrac{n}{n^2}+\dfrac{1}{n^2}}+\dfrac{n}{n}}=\dfrac{1}{1+1}=\dfrac{1}{2}\)
d/ \(=\lim\limits\left[\sqrt{n}\left(\sqrt{3-\dfrac{1}{\sqrt{n}}}-\sqrt{2-\dfrac{1}{\sqrt{n}}}\right)\right]=\lim\limits\left[\sqrt{n}\left(\sqrt{3}-\sqrt{2}\right)\right]=+\infty\)
e/ \(=\lim\limits\dfrac{n^3+2n^2-n-n^3}{\left(\sqrt[3]{n^3+2n^2}\right)^2+n.\sqrt[3]{n^3+2n^2}+n^2}=\lim\limits\dfrac{2n^2-n}{\left(n^3+2n^2\right)^{\dfrac{2}{3}}+n.\left(n^3+2n^2\right)^{\dfrac{1}{3}}+n^2}\)
\(=\dfrac{2}{1+1+1}=\dfrac{2}{3}\)
g/ \(=\lim\limits\dfrac{2^n+9.3^n}{4.3^n+8.2^n}=\lim\limits\dfrac{\left(\dfrac{2}{3}\right)^n+9.\left(\dfrac{3}{3}\right)^n}{4.\left(\dfrac{3}{3}\right)^n+8.\left(\dfrac{2}{3}\right)^n}=\dfrac{9}{4}\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 1 2022
1/...
2/ \(=\lim\dfrac{\dfrac{1}{n\sqrt{n}}-1}{4+\dfrac{1}{n^2\sqrt{n}}}=\dfrac{0-1}{4+0}=-\dfrac{1}{4}\) (chia cả tử-mẫu cho \(n^3\))
3/ \(=\lim\dfrac{3-\left(\dfrac{1}{4}\right)^n}{2.\left(\dfrac{3}{4}\right)^n+4\left(\dfrac{1}{4}\right)^n}=\dfrac{3-0}{2.0+3.0}=\dfrac{3}{0}=+\infty\) (chia tử mẫu cho \(4^n\))
4/ \(=\lim\dfrac{2.2^n+\dfrac{4}{3}.3^n}{1-\dfrac{1}{2}.2^n+3.3^n}=\lim\dfrac{2.\left(\dfrac{2}{3}\right)^n+\dfrac{4}{3}}{\left(\dfrac{1}{3}\right)^n-\dfrac{1}{2}.\left(\dfrac{2}{3}\right)^n+3}=\dfrac{2.0+\dfrac{4}{3}}{0-\dfrac{1}{2}.0+3}=\dfrac{4}{9}\) (chia tử mẫu cho \(3^n\))
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 2 2021
\(a=\lim\dfrac{-2n^2}{\sqrt{n^2+2}+\sqrt{n^2+4}}=\lim\dfrac{-2n}{\sqrt{1+\dfrac{2}{n^2}}+\sqrt{1+\dfrac{4}{n^2}}}=\dfrac{-\infty}{2}=-\infty\)
\(b=\lim\dfrac{3-5n^2+10n}{n-2}=\lim\dfrac{-5n+10+\dfrac{3}{n}}{1-\dfrac{2}{n}}=\dfrac{-\infty}{1}=-\infty\)
\(c=\lim\left(\dfrac{1-\dfrac{1}{n}}{\dfrac{\sqrt{3}}{n}-1}-4.2^n\right)=-1-\infty=-\infty\)
\(d=\lim\dfrac{n^3-4n-\left(3n^2+4\right)\left(n-2\right)}{n^2-2n}=\lim\dfrac{-2n^3+6n^2-8n+8}{n^2-2n}\)
\(\lim\dfrac{-2n+6-\dfrac{8}{n}+\dfrac{8}{n^2}}{1-\dfrac{2}{n}}=\dfrac{-\infty}{1}=-\infty\)
\(e=\lim\dfrac{\sqrt{1+\dfrac{1}{n}}-\sqrt{5}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{5}}=\dfrac{1-\sqrt{5}}{1+\sqrt{5}}\)
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 1 2021
\(a=\lim\left(\dfrac{2n^3\left(5n+1\right)+\left(2n^2+3\right)\left(1-5n^2\right)}{\left(2n^2+3\right)\left(5n+1\right)}\right)\)
\(=\lim\left(\dfrac{2n^3-13n^2+3}{\left(2n^2+3\right)\left(5n+1\right)}\right)=\lim\dfrac{2-\dfrac{13}{n}+\dfrac{3}{n^3}}{\left(2+\dfrac{3}{n^2}\right)\left(5+\dfrac{1}{n}\right)}=\dfrac{2}{2.5}=\dfrac{1}{5}\)
\(b=\lim\left(\dfrac{n-2}{\sqrt{n^2+n}+\sqrt{n^2+2}}\right)=\lim\dfrac{1-\dfrac{2}{n}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1+\dfrac{2}{n}}}=\dfrac{1}{2}\)
\(c=\lim\dfrac{\sqrt{1+\dfrac{3}{n^3}-\dfrac{2}{n^4}}}{2-\dfrac{2}{n}+\dfrac{3}{n^2}}=\dfrac{1}{2}\)
\(d=\lim\dfrac{\sqrt{1-\dfrac{4}{n}}-\sqrt{4+\dfrac{1}{n^2}}}{\sqrt{3+\dfrac{1}{n^2}}-1}=\dfrac{1-2}{\sqrt{3}-1}=-\dfrac{1+\sqrt{3}}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 1 2021
\(a=\lim\sqrt{n^3}\sqrt{\dfrac{1}{n^3}+\dfrac{2}{n^2}-1}=\infty.\left(-1\right)=-\infty\)
\(b=\lim\left(\sqrt{n^2+2n+3}-n+n-\sqrt[3]{n^2+n^3}\right)\)
\(=\lim\dfrac{2n+3}{\sqrt{n^2+2n+3}+n}+\lim\dfrac{-n^2}{n^2+n\sqrt[3]{n^2+n^3}+\sqrt[3]{\left(n^2+n^3\right)^2}}\)
\(=\lim\dfrac{2+\dfrac{3}{n}}{\sqrt{1+\dfrac{2}{n}+\dfrac{3}{n^2}}+1}+\lim\dfrac{-1}{1+\sqrt[3]{\dfrac{1}{n}+1}+\sqrt[3]{\left(\dfrac{1}{n}+1\right)^2}}=\dfrac{2}{2}-\dfrac{1}{3}=\dfrac{2}{3}\)
\(c=\lim\dfrac{\left(\dfrac{2}{\sqrt{n}}+\dfrac{1}{n}\right)\left(\dfrac{1}{\sqrt{n}}+\dfrac{3}{n}\right)}{\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{2}{n}\right)}=\dfrac{0.0}{1.1}=0\)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 1 2021
\(d=\lim\dfrac{4-3\left(\dfrac{2}{4}\right)^n}{9.\left(\dfrac{3}{4}\right)^n+\left(\dfrac{2}{4}\right)^n}=\dfrac{4}{0}=+\infty\)
\(e=\lim\dfrac{7-25\left(\dfrac{5}{7}\right)^n+3.\left(\dfrac{1}{7}\right)^n}{12.\left(\dfrac{6}{7}\right)^n-\left(\dfrac{3}{7}\right)^n+3\left(\dfrac{1}{7}\right)^n}=\dfrac{7}{0}=+\infty\)
\(f=\lim\dfrac{n^4-4n^6}{n\left(\sqrt{n^4+1}+\sqrt{4n^6+1}\right)}=\lim\dfrac{\dfrac{1}{n^2}-6}{\sqrt{\dfrac{1}{n^6}+\dfrac{1}{n^{10}}}+\sqrt{\dfrac{4}{n^4}+\dfrac{1}{n^{10}}}}=\dfrac{-6}{0}=-\infty\)
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 2 2021
\(a=\lim\dfrac{5n\left(n+\sqrt{n^2-n-1}\right)}{n+1}=\lim\dfrac{5\left(n+\sqrt{n^2-n-1}\right)}{1+\dfrac{1}{n}}=\dfrac{+\infty}{1}=+\infty\)
\(b=\lim\dfrac{\sqrt{\dfrac{1}{n}+\sqrt{\dfrac{1}{n^3}+\dfrac{1}{n^4}}}}{1-\dfrac{1}{\sqrt{n}}}=\dfrac{0}{1}=0\)
\(c=\lim\dfrac{\sqrt{2n^2-1+\dfrac{7}{n^2}}}{3+\dfrac{5}{n}}=\dfrac{+\infty}{3}=+\infty\)
\(d=\lim\dfrac{\sqrt{3+\dfrac{2}{n}}-1}{3-\dfrac{2}{n}}=\dfrac{\sqrt{3}-1}{3}\)
HT
14 tháng 1 2021
a/ \(I=lim\dfrac{5^n+2^n}{3^n+4^n}=lim\dfrac{1+\left(\dfrac{2}{5}\right)^n}{\left(\dfrac{3}{5}\right)^n+\left(\dfrac{4}{5}\right)^n}=\dfrac{1}{0}=+\infty\)
b/ \(I=lim\dfrac{\sqrt{n^3+2n}+3n}{n+\sqrt{n^2+1}}=lim\dfrac{\sqrt{\dfrac{n^3}{n^3}+\dfrac{2n}{n^3}}+\dfrac{3n}{n^{\dfrac{3}{2}}}}{\dfrac{n}{n^{\dfrac{3}{2}}}+\sqrt{\dfrac{n^2}{n^3}+\dfrac{1}{n^3}}}=\dfrac{1}{0}=+\infty\)
c/ \(I=lim\left[n\left(\sqrt{2+\dfrac{n}{n^2}}-\sqrt{1+\dfrac{2n}{n^2}+\dfrac{3}{n^2}}\right)\right]=+\infty.\left(\sqrt{2}-1\right)=+\infty\)
AH
Akai Haruma
Giáo viên
29 tháng 1 2023
a.
\(A=\lim\frac{\sqrt[3]{n^6-7n^3-5n+8}}{n+12}=\lim \frac{\sqrt[3]{\frac{n^6-7n^3-5n+8}{n^3}}}{\frac{n+12}{n}}=\lim \frac{\sqrt[3]{n^3-7-\frac{5}{n^2}+\frac{8}{n^3}}}{1+\frac{12}{n}}\)
Ta thấy:
\(\lim\sqrt[3]{n^3-7-\frac{5}{n^2}+\frac{8}{n^3}}=\infty \)
\(\lim (1+\frac{12}{n})=1\)
Suy ra $A=\infty$
AH
Akai Haruma
Giáo viên
29 tháng 1 2023
b.
\(B=\lim\frac{1}{\sqrt{3n+2}-\sqrt{2n+1}}=\lim \frac{1}{\frac{3n+2-(2n+1)}{\sqrt{3n+2}+\sqrt{2n+1}}}=\lim \frac{\sqrt{3n+2}+\sqrt{2n+1}}{n+1}\)
\(=\lim \frac{\sqrt{\frac{3n+2}{n}}+\sqrt{\frac{2n+1}{n}}}{\frac{n+1}{\sqrt{n}}}=\lim \frac{\sqrt{3+\frac{2}{n}}+\sqrt{2+\frac{1}{n}}}{\sqrt{n}+\frac{1}{\sqrt{n}}}\)
Ta thấy:
\(\lim( \sqrt{3+\frac{2}{n}}+\sqrt{2+\frac{1}{n}})=\sqrt{3}+\sqrt{2}>0\)
\(\lim (\sqrt{n}+\frac{1}{\sqrt{n}})=\infty\)
$\Rightarrow B=\infty$
\(lim\dfrac{n^2+\sqrt[3]{1-n^6}}{\sqrt{n^4+1}-n^2}=lim\dfrac{\left(n^2+\sqrt[3]{1-n^6}\right)\left(n^2+\sqrt{n^4+1}\right)}{n^4+1-n^4}\)
\(=lim\left(n^2+\sqrt[3]{n^6\left(\dfrac{1}{n^6}-1\right)}\right)\left(n^2+\sqrt{n^4\left(1+\dfrac{1}{n^4}\right)}\right)\)
\(=lim\left(n^2-n^2\right)\left(n^2+n^2\right)=0\)