Chứng minh rằng :
\(\lim\limits_{n\rightarrow+\infty}\int\limits^1_0x^n\sin\pi xdx=0\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
14 tháng 10 2023
1) \(\lim\limits_{n\rightarrow\infty}\dfrac{3n^2+5n-3}{-n+5}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n\left(3n+5-\dfrac{3}{n}\right)}{-n\left(1-\dfrac{5}{n}\right)}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3n+5-\dfrac{3}{n}}{-\left(1-\dfrac{5}{n}\right)}\)
\(=\left[{}\begin{matrix}-\infty\left(n\rightarrow+\infty\right)\\+\infty\left(n\rightarrow-\infty\right)\end{matrix}\right.\)
Bài 2,3 tương tự, bạn tự làm nhé!
MH
13 tháng 10 2023
1) \(\lim\limits_{n\rightarrow\infty}\dfrac{6n-8}{n-1}=\lim\limits_{n\rightarrow\infty}\dfrac{2n\left(1-\dfrac{4}{n}\right)}{n\left(1-\dfrac{1}{n}\right)}=2\)
2) \(\lim\limits_{n\rightarrow\infty}\dfrac{n^2+5n-3}{4n^3-2n+5}=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(1+\dfrac{5}{n}-\dfrac{3}{n^2}\right)}{n^3\left(4-\dfrac{2}{n^2}+\dfrac{5}{n^3}\right)}=\dfrac{1}{4n}=\infty\)
3) \(\lim\limits_{n\rightarrow\infty}\left(-2n^5+4n^4-3n^2+4\right)=\lim\limits_{n\rightarrow\infty}n^5\left(-2+\dfrac{4}{n}-\dfrac{3}{n^2}+\dfrac{4}{n^5}\right)=-2n^5=-\infty\)
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.
15 tháng 10 2023
3:
\(\lim\limits_{n\rightarrow\infty}\dfrac{2-5^{n-2}}{3^n+2\cdot5^n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{2}{5^n}-\dfrac{5^{n-2}}{5^n}}{\dfrac{3^n}{5^n}+2\cdot\dfrac{5^n}{5^n}}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{2}{5^n}-\dfrac{1}{25}}{\left(\dfrac{3}{5}\right)^n+2\cdot1}\)
\(=-\dfrac{1}{25}:2=-\dfrac{1}{50}\)
1:
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3^n-4^n\cdot4}{3^n\cdot9+4^n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{3^n}{4^n}-4}{3^n\cdot\dfrac{9}{4^n}+1}\)
\(=-\dfrac{4}{1}=-4\)
15 tháng 10 2023
1: \(\lim\limits_{n->\infty}\dfrac{2n+1}{n+15}=\lim\limits_{n\rightarrow\infty}\dfrac{2+\dfrac{1}{n}}{1+\dfrac{15}{n}}=2\)
2: \(\lim\limits_{n\rightarrow\infty}\dfrac{n+6}{2n-5}=\lim\limits_{n\rightarrow\infty}\dfrac{1+\dfrac{6}{n}}{2-\dfrac{5}{n}}=\dfrac{1}{2}\)
15 tháng 10 2023
1: \(\lim\limits_{n\rightarrow\infty}\dfrac{-7n^2+4}{-n+5}=\lim\limits_{n\rightarrow\infty}\dfrac{7n^2-4}{n-5}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(7-\dfrac{4}{n^2}\right)}{n\left(1-\dfrac{5}{n}\right)}=\lim\limits_{n\rightarrow\infty}\dfrac{n\left(7-\dfrac{4}{n^2}\right)}{1-\dfrac{5}{n}}\)
\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{n\rightarrow\infty}n=+\infty\\\lim\limits_{n\rightarrow\infty}\dfrac{7-\dfrac{4}{n^2}}{1-\dfrac{5}{n}}=\dfrac{7}{1}=7>0\end{matrix}\right.\)
2:
\(\lim\limits_{n\rightarrow\infty}\dfrac{-3n^2+2}{n-2}=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(-3+\dfrac{2}{n^2}\right)}{n\left(1-\dfrac{2}{n}\right)}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n\left(-3+\dfrac{2}{n^2}\right)}{1-\dfrac{2}{n}}\)
\(=-\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{n\rightarrow\infty}n=+\infty\\\lim\limits_{n\rightarrow\infty}\dfrac{-3+\dfrac{2}{n^2}}{1-\dfrac{2}{n}}=-\dfrac{3}{1}=-3< 0\end{matrix}\right.\)