Tính lim(\(\sqrt{4n^2+2}.\sqrt[3]{n^3+1}-2n\sqrt[3]{n^3+2}\))
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NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 1 2021
\(a=\lim\left(\dfrac{2n^3\left(5n+1\right)+\left(2n^2+3\right)\left(1-5n^2\right)}{\left(2n^2+3\right)\left(5n+1\right)}\right)\)
\(=\lim\left(\dfrac{2n^3-13n^2+3}{\left(2n^2+3\right)\left(5n+1\right)}\right)=\lim\dfrac{2-\dfrac{13}{n}+\dfrac{3}{n^3}}{\left(2+\dfrac{3}{n^2}\right)\left(5+\dfrac{1}{n}\right)}=\dfrac{2}{2.5}=\dfrac{1}{5}\)
\(b=\lim\left(\dfrac{n-2}{\sqrt{n^2+n}+\sqrt{n^2+2}}\right)=\lim\dfrac{1-\dfrac{2}{n}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1+\dfrac{2}{n}}}=\dfrac{1}{2}\)
\(c=\lim\dfrac{\sqrt{1+\dfrac{3}{n^3}-\dfrac{2}{n^4}}}{2-\dfrac{2}{n}+\dfrac{3}{n^2}}=\dfrac{1}{2}\)
\(d=\lim\dfrac{\sqrt{1-\dfrac{4}{n}}-\sqrt{4+\dfrac{1}{n^2}}}{\sqrt{3+\dfrac{1}{n^2}}-1}=\dfrac{1-2}{\sqrt{3}-1}=-\dfrac{1+\sqrt{3}}{2}\)
HT
16 tháng 2 2021
a/ Bạn coi lại đề bài, 3n^2 +n^2 thì bằng 4n^2 luôn chứ ko ai cho đề bài như vậy cả
b/ \(\lim\limits\dfrac{\dfrac{n^3}{n^3}+\dfrac{3n}{n^3}+\dfrac{1}{n^3}}{-\dfrac{n^3}{n^3}+\dfrac{2n}{n^3}}=-1\)
c/ \(=\lim\limits\dfrac{-\dfrac{2n^3}{n^2}+\dfrac{3n}{n^2}+\dfrac{1}{n^2}}{-\dfrac{n^2}{n^2}+\dfrac{n}{n^2}}=\lim\limits\dfrac{-2n}{-1}=+\infty\)
d/ \(=\lim\limits\left[n\left(1+1\right)\right]=+\infty\)
e/ \(\lim\limits\left[2^n\left(\dfrac{2n}{2^n}-3+\dfrac{1}{2^n}\right)\right]=\lim\limits\left(-3.2^n\right)=-\infty\)
f/ \(=\lim\limits\dfrac{4n^2-n-4n^2}{\sqrt{4n^2-n}+2n}=\lim\limits\dfrac{-\dfrac{n}{n}}{\sqrt{\dfrac{4n^2}{n^2}-\dfrac{n}{n^2}}+\dfrac{2n}{n}}=-\dfrac{1}{2+2}=-\dfrac{1}{4}\)
g/ \(=\lim\limits\dfrac{n^2+3n-1-n^2}{\sqrt{n^2+3n-1}+n}+\lim\limits\dfrac{n^3-n^3+n}{\sqrt[3]{\left(n^3-n\right)^2}+n.\sqrt[3]{n^3-n}+n^2}\)
\(=\lim\limits\dfrac{\dfrac{3n}{n}-\dfrac{1}{n}}{\sqrt{\dfrac{n^2}{n^2}+\dfrac{3n}{n^2}-\dfrac{1}{n^2}}+\dfrac{n}{n}}+\lim\limits\dfrac{\dfrac{n}{n^2}}{\dfrac{\sqrt[3]{\left(n^3-n\right)^2}}{n^2}+\dfrac{n\sqrt[3]{n^3-n}}{n^2}+\dfrac{n^2}{n^2}}\)
\(=\dfrac{3}{2}+0=\dfrac{3}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 2 2021
\(a=\lim\dfrac{1}{\sqrt{4n+1}+2\sqrt{n}}=\dfrac{1}{\infty}=0\)
\(b=\lim n\left(\sqrt{1+\dfrac{2}{n}}-\sqrt{1-\dfrac{2}{n}}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(c=\lim4^n\left(\sqrt{\left(\dfrac{9}{16}\right)^n-\left(\dfrac{3}{16}\right)^n}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(d=\lim n^3\left(3+\dfrac{2}{n}+\dfrac{1}{n^2}\right)=+\infty.3=+\infty\)
BD
1
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 1 2021
\(=\lim\dfrac{\sqrt{4-\dfrac{1}{n}}+\sqrt[3]{8+\dfrac{1}{n}}}{2+\dfrac{3}{n}}=\dfrac{2+2}{2}=2\)
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/ \(lim\left(\frac{6.6^n+3^5.9^n}{3^3.9^n-\frac{1}{2}.4^n}\right)=lim\left(\frac{6\left(\frac{2}{3}\right)^n+3^5}{3^3-\frac{1}{2}\left(\frac{4}{9}\right)^n}\right)=\frac{3^5}{3^3}=9\)
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\(\frac{2+\frac{3}{n}}{\sqrt{9+\frac{3}{n^2}}-\sqrt[3]{\frac{2}{n}-8}}=\frac{2}{3+2}=\frac{2}{5}\)
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.
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 \(\frac{\sqrt[3]{8+\frac{1}{n^3}}-1}{2-\frac{3}{n}}=\frac{2-1}{2}=\frac{1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 1 2019
\(lim\dfrac{\left(2-n\right)\left(3+2n^3\right)}{2n^2-1}=lim\dfrac{\left(\dfrac{2}{n}-1\right)\left(\dfrac{3}{n}+2n^2\right)}{2-\dfrac{1}{n^2}}=-\infty\)
\(\dfrac{lim\left(\sqrt{4n^2+1}-2n\right)n}{\sqrt[3]{4-n^3}+n}=lim\dfrac{n\left(\sqrt[3]{\left(4-n^3\right)^2}-n\sqrt[3]{4-n^3}+n^2\right)}{4.\left(\sqrt{4n^2+1}+2n\right)}\)
\(=lim\dfrac{\sqrt[3]{\left(n^3-4\right)^2}+n\sqrt[3]{n^3-4}+n^2}{4\left(\sqrt{4+\dfrac{1}{n^2}}+2\right)}=+\infty\)
\(lim\left(\sqrt{4n^2+2}\sqrt[3]{n^3+1}-2n\sqrt[3]{n^3+2}\right)\\ =lim\left[\sqrt[3]{n^3+1}\left(\sqrt{4n^2+2}-2n\right)-2n\left(\sqrt[3]{n^3+2}-\sqrt[3]{n^3+1}\right)\right]\\ =lim\left[\dfrac{2\sqrt[3]{n^3+1}}{\sqrt{4n^2+2}+2n}-\dfrac{2n}{\left(\sqrt[3]{n^3+2}\right)^2+\left(\sqrt[3]{n^3+1}\right)^2+\sqrt[3]{n^3+2}\sqrt[3]{n^3+1}}\right]\\ =lim\left[\dfrac{2\sqrt[3]{1+\dfrac{1}{n^3}}}{\sqrt{4+\dfrac{2}{n^2}}+2}-\dfrac{\dfrac{2}{n}}{\left(\sqrt[3]{1+\dfrac{2}{n}}\right)^2+\left(\sqrt[3]{1+\dfrac{1}{n}}\right)^2+\sqrt[3]{1+\dfrac{2}{n}}\sqrt[3]{1+\dfrac{1}{n}}}\right]\)
\(=\dfrac{1}{2}\)