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Cấp số nhân lùi vô hạn có \(u_1=1\) và \(q=\dfrac{1}{2}\)
\(\Rightarrow S=\dfrac{u_1}{1-q}=\dfrac{1}{1-\dfrac{1}{2}}=2\)
a: \(S=\dfrac{\dfrac{2}{3}}{\dfrac{5}{4}}=\dfrac{8}{15}\)
b: 1,(6)=5/3
\(S=1+\dfrac{1}{2}+...+\dfrac{1}{2^{n-1}}\) là tổng cấp số nhân với \(\left\{{}\begin{matrix}u_1=1\\q=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow S=1.\dfrac{1-\left(\dfrac{1}{2}\right)^n}{1-\dfrac{1}{2}}=2-\dfrac{1}{2^{n-1}}\)
Do đó: \(\lim\left(S\right)=\lim\left(2-\dfrac{1}{2^{n-1}}\right)=2\)
\(=lim\dfrac{\left(1-\dfrac{1}{3^{n-1}}\right)\left(1-\dfrac{2}{5}\right)}{\left(1-\dfrac{1}{3}\right)\left(1-\left(\dfrac{2}{50}\right)^{n+1}\right)}\\ =lim\dfrac{9}{10}\left(\dfrac{1-\dfrac{1}{3^{n-1}}}{1-\left(\dfrac{-2}{5}\right)^{n+1}}\right)\\ =\dfrac{9}{10}\)
\(a=lim\dfrac{\left(\dfrac{2}{6}\right)^n+1-\dfrac{1}{4}\left(\dfrac{4}{6}\right)^n}{\left(\dfrac{3}{6}\right)^n+6}=\dfrac{1}{6}\)
\(b=\lim\dfrac{\left(n+1\right)^2}{3n^2+4}=\lim\dfrac{n^2+2n+1}{3n^2+4}=\lim\dfrac{1+\dfrac{2}{n}+\dfrac{1}{n^2}}{3+\dfrac{4}{n^2}}=\dfrac{1}{3}\)
\(c=\lim\dfrac{n\left(n+1\right)}{2\left(n^2-3\right)}=\lim\dfrac{n^2+n}{2n^2-6}=\lim\dfrac{1+\dfrac{1}{n}}{2-\dfrac{6}{n^2}}=\dfrac{1}{2}\)
\(d=\lim\left[1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right]=\lim\left[1-\dfrac{1}{n+1}\right]=1\)
\(e=\lim\dfrac{1}{2}\left[1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right]\)
\(=\lim\dfrac{1}{2}\left[1-\dfrac{1}{2n+1}\right]=\dfrac{1}{2}\)
1:
\(\lim\limits_{n\rightarrow\infty}\dfrac{7^n+4}{3\cdot7^n+4^n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{1+\dfrac{4}{7^n}}{3+\left(\dfrac{4}{7}\right)^n}=\dfrac{1}{3}\)
2: \(\lim\limits_{n\rightarrow\infty}\dfrac{1-4^n}{1+4^n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{1}{4^n}-1}{\dfrac{1}{4^n}+1}=-\dfrac{1}{1}=-1\)
\(1,\lim\limits_{n\rightarrow\infty}\dfrac{-n^2+2n+1}{\sqrt{3n^4+2}}\left(1\right)\)
\(\dfrac{-n^2+2n+1}{\sqrt{3n^4+2}}=\dfrac{-\dfrac{n^2}{n^4}+\dfrac{2n}{n^4}+\dfrac{1}{n^4}}{\sqrt{\dfrac{3n^4}{n^4}+\dfrac{2}{n^4}}}=\dfrac{-\dfrac{1}{n^2}+\dfrac{2}{n^3}+\dfrac{1}{n^4}}{\sqrt{3+\dfrac{2}{n^4}}}\)
\(\Rightarrow\left(1\right)=\dfrac{-lim\dfrac{1}{n^2}+2lim\dfrac{1}{n^3}+lim\dfrac{1}{n^4}}{\sqrt{lim\left(3+\dfrac{2}{n^4}\right)}}\)
\(=\dfrac{0}{\sqrt{lim\left(3+\dfrac{2}{n^4}\right)}}=0\)
\(2,\lim\limits_{n\rightarrow\infty}\left(\dfrac{4n-\sqrt{16n^2+1}}{n+1}\right)\left(2\right)\)
\(\dfrac{4n-\sqrt{16n^2+1}}{n+1}=\dfrac{\dfrac{4n}{n^2}-\sqrt{\dfrac{16n^2}{n^2}+\dfrac{1}{n^2}}}{\dfrac{n}{n^2}+\dfrac{1}{n^2}}=\dfrac{\dfrac{4}{n}-\sqrt{16+\dfrac{1}{n^2}}}{\dfrac{1}{n}+\dfrac{1}{n^2}}\)
\(\Rightarrow\left(2\right)=\dfrac{lim\left(\dfrac{4}{n}-\sqrt{16+\dfrac{1}{n^2}}\right)}{lim\left(\dfrac{1}{n}+\dfrac{1}{n^2}\right)}=\dfrac{lim\left(\dfrac{4}{n}-\sqrt{16+\dfrac{1}{n^2}}\right)}{0}\)
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.
2:
\(\lim\limits_{n\rightarrow\infty}\dfrac{3^n+1}{2^n-1}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{3^n}{3^n}+\dfrac{1}{3^n}}{\dfrac{2^n}{3^n}-\dfrac{1}{3^n}}=\lim\limits_{n\rightarrow\infty}\dfrac{1+\dfrac{1}{3^n}}{\left(\dfrac{2}{3}\right)^n-\dfrac{1}{3^n}}=1\)
\(\left\{{}\begin{matrix}u_1=\dfrac{1}{2}\\q=-\dfrac{1}{2}\end{matrix}\right.\Rightarrow S=\dfrac{1}{2}.\dfrac{1}{1+\dfrac{1}{2}}=\dfrac{1}{3}\)