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PH
25 tháng 2 2020
lim\(\frac{3n^2+n-5}{2n^2+1}\)=lim\(\frac{n^2\left(3+\frac{1}{n}-\frac{5}{n^2}\right)}{n^2\left(2+\frac{1}{n}\right)}\)=\(\frac{3}{2}\)
PH
25 tháng 2 2020
lim\(\frac{\sqrt{9n^2-n}+1}{4n-2}\)=lim\(\frac{n\sqrt{9-\frac{1}{n}+\frac{1}{n^2}}}{n\left(4-\frac{2}{n}\right)}\)=lim\(\frac{\sqrt{9}}{4}\)=\(\frac{3}{2}\)
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\(\frac{2+\frac{3}{n}}{\sqrt{9+\frac{3}{n^2}}-\sqrt[3]{\frac{2}{n}-8}}=\frac{2}{3+2}=\frac{2}{5}\)
HM
1
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 1 2019
\(lim\dfrac{5n\sqrt{2n^2-n}}{1+5n-3n^2}=lim\dfrac{5\sqrt{2-\dfrac{1}{n}}}{\dfrac{1}{n^2}+\dfrac{5}{n}-3}=\dfrac{5\sqrt{2-0}}{0+0-3}=\dfrac{-5\sqrt{2}}{3}\)
\(lim\dfrac{\sqrt{4n^2+n}-7n}{3n^2-1}=lim\dfrac{\sqrt{\dfrac{4}{n^2}+\dfrac{1}{n^3}}-\dfrac{7}{n}}{3-\dfrac{1}{n^2}}=\dfrac{\sqrt{0+0}-0}{3-0}=\dfrac{0}{3}=0\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 1 2019
\(lim\dfrac{\left(2-n\right)\left(3+2n^3\right)}{2n^2-1}=lim\dfrac{\left(\dfrac{2}{n}-1\right)\left(\dfrac{3}{n}+2n^2\right)}{2-\dfrac{1}{n^2}}=-\infty\)
\(\dfrac{lim\left(\sqrt{4n^2+1}-2n\right)n}{\sqrt[3]{4-n^3}+n}=lim\dfrac{n\left(\sqrt[3]{\left(4-n^3\right)^2}-n\sqrt[3]{4-n^3}+n^2\right)}{4.\left(\sqrt{4n^2+1}+2n\right)}\)
\(=lim\dfrac{\sqrt[3]{\left(n^3-4\right)^2}+n\sqrt[3]{n^3-4}+n^2}{4\left(\sqrt{4+\dfrac{1}{n^2}}+2\right)}=+\infty\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 1 2022
7/
\(=\lim\dfrac{n^2+4n+1-n^2}{\sqrt{n^2+4n+1}+n}=\lim\dfrac{4n+1}{\sqrt{n^2+4n+1}+n}=\lim\dfrac{4+\dfrac{1}{n}}{\sqrt{1+\dfrac{4}{n}+\dfrac{1}{n^2}}+1}=\dfrac{4}{1+1}=2\)
8/
\(=\lim\dfrac{n^2-\left(n^2+9n-1\right)}{n+\sqrt{n^2+9n-1}}=\lim\dfrac{-9n+1}{n+\sqrt{n^2+9n-1}}=\lim\dfrac{-9+\dfrac{1}{n}}{1+\sqrt{1+\dfrac{9}{n}-\dfrac{1}{n^2}}}=\dfrac{-9}{1+1}=-\dfrac{9}{2}\)
9/
Do \(1+2+...+n=\dfrac{n\left(n+1\right)}{2}=\dfrac{n^2+n}{2}\)
\(\Rightarrow\lim\dfrac{1+2+...+n}{n^2-1}=\lim\dfrac{n^2+n}{2n^2-2}=\lim\dfrac{1+\dfrac{1}{n}}{2-\dfrac{2}{n^2}}=\dfrac{1}{2}\)
DT
1
28 tháng 3 2020
\(lim\frac{\sqrt{4n^2+1}+2n-1}{\sqrt{n^2+4n+1}+n}\)
= \(lim\frac{\sqrt{4+\frac{1}{n^2}}+2-\frac{1}{n}}{\sqrt{1+\frac{4}{n}+\frac{1}{n^2}}+1}\)
=\(\frac{2+2}{1+1}=2\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 2 2020
a/ \(lim\left(\sqrt[3]{n-n^3}+n+\sqrt{n^2+3n}-n\right)\)
\(=lim\left(\frac{n}{\sqrt[3]{\left(n-n^3\right)^2}-n\sqrt[3]{\left(n-n^3\right)}+n^2}+\frac{3n}{\sqrt{n^2+3n}+n}\right)\)
\(=lim\left(\frac{1}{\sqrt[3]{n^3+2n+\frac{1}{n}}+\sqrt[3]{n^3-n}+n}+\frac{3}{\sqrt{1+\frac{3}{n}}+1}\right)=0+\frac{3}{1+1}=\frac{3}{2}\)
b/ \(lim\left(\frac{-2\sqrt{n}-4}{\sqrt{n-2\sqrt{n}}+\sqrt{n+4}}\right)=lim\left(\frac{-2-\frac{4}{\sqrt{n}}}{\sqrt{1-\frac{2}{\sqrt{n}}}+\sqrt{1+\frac{4}{n}}}\right)=-\frac{2}{1+1}=-1\)
c/ \(lim\left(\frac{3n^2}{\sqrt[3]{n^6+6n^5+9n^4}+\sqrt[3]{n^6+3n^5}+n^2}\right)=lim\left(\frac{3}{\sqrt[3]{1+\frac{6}{n}+\frac{9}{n^2}}+\sqrt[3]{1+\frac{3}{n}}+1}\right)=\frac{3}{3}=1\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 2 2020
d/ \(lim\left(\sqrt[3]{n^3+6n}-n+n-\sqrt{n^2-4n}\right)=lim\left(\frac{6n}{\sqrt[3]{n^6+12n^4+36n^2}+\sqrt[3]{n^6+6n^4}+n^2}+\frac{4n}{n+\sqrt{n^2-4n}}\right)\)
\(=lim\left(\frac{6}{\sqrt[3]{n^3+12n+\frac{36}{n}}+\sqrt[3]{n^3+6n}+n}+\frac{4}{1+\sqrt{1-\frac{4}{n}}}\right)=0+\frac{4}{1+1}=2\)
e/ \(lim\left(\frac{-3.3^n+4.4^n}{5.3^n+\frac{3}{2}.4^n}\right)=lim\left(\frac{-3\left(\frac{3}{4}\right)^n+4}{5.\left(\frac{3}{4}\right)^n+\frac{3}{2}}\right)=\frac{0+4}{0+\frac{3}{2}}=\frac{8}{3}\)
f/ \(lim\left(\frac{9^n-5.5^n+7.7^n}{9.3^n+5^n+2.8^n}\right)=lim\left(\frac{1-5.\left(\frac{5}{9}\right)^n+7\left(\frac{7}{9}\right)^n}{9.\left(\frac{1}{3}\right)^n+\left(\frac{5}{9}\right)^n+2.\left(\frac{8}{9}\right)^n}\right)=\frac{1}{0}=+\infty\)
g/ \(lim\left(\frac{6.6^n+3^5.9^n}{3^3.9^n-\frac{1}{2}.4^n}\right)=lim\left(\frac{6\left(\frac{2}{3}\right)^n+3^5}{3^3-\frac{1}{2}\left(\frac{4}{9}\right)^n}\right)=\frac{3^5}{3^3}=9\)
1 tháng 4 2020
a)lim \(\frac{\sqrt{n^2-4n}-\sqrt{4n+1}}{\sqrt{3n^2+1}+n}\)
=lim \(\frac{\sqrt{1-\frac{4}{n}}-\sqrt{\frac{4}{n}+\frac{1}{n^2}}}{\sqrt{3+\frac{1}{n^2}}+1}=\frac{1}{\sqrt{3}+1}\)
b)lim \(\frac{\sqrt[3]{8n^3+n^2}-n}{2n-3}\)
= lim \(\frac{\sqrt[3]{8+\frac{1}{n^3}}-1}{2-\frac{3}{n}}=\frac{2-1}{2}=\frac{1}{2}\)
=3/4
3/4 nha bạn