\(\left(\dfrac{1}{3}\right)^{2n-1}=3^5\)
Cho:
\(A=\dfrac{1}{1.\left(2n-1\right)}+\dfrac{1}{3.\left(2n-3\right)}+...+\dfrac{1}{\left(2n-3\right).3}+\dfrac{1}{\left(2n-1\right).1}\) \(B=1+\dfrac{1}{3}+...+\dfrac{1}{2n-1}\) (với n ∈ N*).
Tính \(\dfrac{A}{B}\)
Cho A = \(\dfrac{1}{1.\left(2n-1\right)}+\dfrac{1}{3.\left(2n-3\right)}+...+\dfrac{1}{3.\left(2n-3\right)}+\dfrac{1}{1.\left(2n-1\right)}\); B = \(1+\dfrac{1}{3}+...+\dfrac{1}{2n-1}\). Tính \(\dfrac{A}{B}\)
cho
A=\(\dfrac{1}{1\left(2n-1\right)}+\dfrac{1}{3\left(2n-3\right)}+...+\dfrac{1}{\left(2n-3\right)3}+\dfrac{1}{\left(2n-1\right)1}\)
B=\(1+\dfrac{1}{3}+...+\dfrac{1}{2n-1}\)
tính \(\dfrac{A}{B}\)
Tìm các giới hạn sau:
a) \(lim\sqrt[3]{-n^3+2n^2-5}\)
b) \(lim\dfrac{1}{\sqrt{n+1}-\sqrt{n}}\)
c) \(lim\left(\dfrac{1}{n+1}-n\right)\)
d) \(lim\left(\dfrac{2n^2-1}{n+1}-2n\right)\)
e) \(lim\dfrac{2n^3+n^2-3n+1}{2-3n}\)
\(a=\lim n\left(\sqrt[3]{-1+\dfrac{2}{n}-\dfrac{5}{n^3}}\right)=+\infty.\left(-1\right)=-\infty\)
\(b=\lim\left(\sqrt{n+1}+\sqrt{n}\right)=+\infty\)
\(c=\lim n\left(\dfrac{1}{n^2+n}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(d=\lim\left(\dfrac{2n^2-1-2n\left(n+1\right)}{n+1}\right)=\lim\left(\dfrac{-1-2n}{n+1}\right)=-2\)
\(e=\lim\dfrac{2n^2+n-3+\dfrac{1}{n}}{\dfrac{2}{n}-3}=\dfrac{+\infty}{-3}=-\infty\)
\(lim\dfrac{4n^5+n^3-5n+3}{\left(2n^2-3\right)\left(2n+1\right)^3}\)
\(lim\dfrac{4n^5+n^3-5n+3}{\left(2n^2-3\right)\left(2n+1\right)^3}\)
Lời giải:
\(\lim \frac{4n^5+n^3-5n+3}{(2n^2-3)(2n+1)^3}=\lim \frac{4+\frac{1}{n^2}-\frac{5}{n^4}+\frac{3}{n^5}}{(2-\frac{3}{n^2})(2+\frac{1}{n})^3}=\frac{4}{2.2^3}=\frac{1}{4}\)
1
a,Lim\(\sqrt{1+2n-n^3}\)
b,Lim\(\sqrt{n^2+2n+3}-\sqrt[3]{n^2+n^3}\)
c,Lim\(\dfrac{\left(2\sqrt{n}+1\right)\left(\sqrt{n}+3\right)}{\left(n+1\right)\left(n+2\right)}\)
d,\(\dfrac{4^{n+1}-3\times2^n}{3^{n+2}+2^n}\)
e,\(\dfrac{7^{n+1}-5^{n+2}+3}{2\times6^{n+1}-3^n+3}\)
f,\(\dfrac{\sqrt{n^4+1}}{n}\) -\(\dfrac{\sqrt{4n^6+1}}{n}\)
\(a=\lim\sqrt{n^3}\sqrt{\dfrac{1}{n^3}+\dfrac{2}{n^2}-1}=\infty.\left(-1\right)=-\infty\)
\(b=\lim\left(\sqrt{n^2+2n+3}-n+n-\sqrt[3]{n^2+n^3}\right)\)
\(=\lim\dfrac{2n+3}{\sqrt{n^2+2n+3}+n}+\lim\dfrac{-n^2}{n^2+n\sqrt[3]{n^2+n^3}+\sqrt[3]{\left(n^2+n^3\right)^2}}\)
\(=\lim\dfrac{2+\dfrac{3}{n}}{\sqrt{1+\dfrac{2}{n}+\dfrac{3}{n^2}}+1}+\lim\dfrac{-1}{1+\sqrt[3]{\dfrac{1}{n}+1}+\sqrt[3]{\left(\dfrac{1}{n}+1\right)^2}}=\dfrac{2}{2}-\dfrac{1}{3}=\dfrac{2}{3}\)
\(c=\lim\dfrac{\left(\dfrac{2}{\sqrt{n}}+\dfrac{1}{n}\right)\left(\dfrac{1}{\sqrt{n}}+\dfrac{3}{n}\right)}{\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{2}{n}\right)}=\dfrac{0.0}{1.1}=0\)
\(d=\lim\dfrac{4-3\left(\dfrac{2}{4}\right)^n}{9.\left(\dfrac{3}{4}\right)^n+\left(\dfrac{2}{4}\right)^n}=\dfrac{4}{0}=+\infty\)
\(e=\lim\dfrac{7-25\left(\dfrac{5}{7}\right)^n+3.\left(\dfrac{1}{7}\right)^n}{12.\left(\dfrac{6}{7}\right)^n-\left(\dfrac{3}{7}\right)^n+3\left(\dfrac{1}{7}\right)^n}=\dfrac{7}{0}=+\infty\)
\(f=\lim\dfrac{n^4-4n^6}{n\left(\sqrt{n^4+1}+\sqrt{4n^6+1}\right)}=\lim\dfrac{\dfrac{1}{n^2}-6}{\sqrt{\dfrac{1}{n^6}+\dfrac{1}{n^{10}}}+\sqrt{\dfrac{4}{n^4}+\dfrac{1}{n^{10}}}}=\dfrac{-6}{0}=-\infty\)
Cho \(S_n=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\). Kết quả giới hạn của limSn bằng:
A. \(\dfrac{1}{3}\)
B. \(\dfrac{1}{5}\)
C. \(\dfrac{1}{2}\)
D. 0
\(S_n=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)=\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{2n}{2n+1}\right)=\dfrac{2n}{2\left(2n+1\right)}\)
-> chọn D
Tính :6/ lim\(\dfrac{-n^2+2n+1}{\sqrt{3n^4+2}}\)
7/ lim \(\dfrac{\sqrt{n^3-2n+5}}{3+5n}\)
10/ lim\(\dfrac{1+3+5+...+\left(2n+1\right)}{3n^3+4}\)
Tìm giới hạn dãy số sau
\(lim\dfrac{\left(2n-1\right)\left(3n^2+2\right)^3}{-2n^5+4n^3-1}\)
\(lim\left(3.2^{n+1}-5.3^n+7n\right)\)
\(\lim\dfrac{\left(2n-1\right)\left(3n^2+2\right)^3}{-2n^5+4n^3-1}=\lim\dfrac{\left(\dfrac{2n-1}{n}\right)\left(\dfrac{3n^2+2}{n^2}\right)^3}{\dfrac{-2n^5+4n^3-1}{n^7}}\)
\(=\lim\dfrac{\left(2-\dfrac{1}{n}\right)\left(3+\dfrac{2}{n^2}\right)^3}{-\dfrac{2}{n^2}+\dfrac{4}{n^4}-\dfrac{1}{n^7}}=-\infty\)
\(\lim3^n\left(6.\left(\dfrac{2}{3}\right)^n-5+\dfrac{7n}{3^n}\right)=+\infty.\left(-5\right)=-\infty\)