Chứng minh các giới hạn sau :
\(1.lim\dfrac{n^2+1}{n}=+\infty\)
\(2.lim\dfrac{2-n}{\sqrt{n}}=-\infty\)
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
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:
\(\lim\limits_{n\rightarrow\infty}\dfrac{7^n+4}{3\cdot7^n+4^n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{1+\dfrac{4}{7^n}}{3+\left(\dfrac{4}{7}\right)^n}=\dfrac{1}{3}\)
2: \(\lim\limits_{n\rightarrow\infty}\dfrac{1-4^n}{1+4^n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{1}{4^n}-1}{\dfrac{1}{4^n}+1}=-\dfrac{1}{1}=-1\)
15 tháng 10 2023
2:
\(\lim\limits_{n\rightarrow\infty}\dfrac{3^n+1}{2^n-1}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{3^n}{3^n}+\dfrac{1}{3^n}}{\dfrac{2^n}{3^n}-\dfrac{1}{3^n}}=\lim\limits_{n\rightarrow\infty}\dfrac{1+\dfrac{1}{3^n}}{\left(\dfrac{2}{3}\right)^n-\dfrac{1}{3^n}}=1\)
AH
Akai Haruma
Giáo viên
30 tháng 11 2023
1.
Trước hết bạn nhớ công thức:
$1^2+2^2+....+n^2=\frac{n(n+1)(2n+1)}{6}$ (cách cm ở đây: https://hoc24.vn/cau-hoi/tinh-tongs-122232n2.83618073020)
Áp vào bài:
\(\lim\frac{1}{n^3}[1^2+2^2+....+(n-1)^2]=\lim \frac{1}{n^3}.\frac{(n-1)n(2n-1)}{6}=\lim \frac{n(n-1)(2n-1)}{6n^3}\)
\(=\lim \frac{(n-1)(2n-1)}{6n^2}=\lim (\frac{n-1}{n}.\frac{2n-1}{6n})=\lim (1-\frac{1}{n})(\frac{1}{3}-\frac{1}{6n})\)
\(=1.\frac{1}{3}=\frac{1}{3}\)
AH
Akai Haruma
Giáo viên
30 tháng 11 2023
2.
\(\lim \frac{1}{n}\left[(x+\frac{a}{n})+(x+\frac{2a}{n})+...+(x.\frac{(n-1)a}{n}\right]\)
\(=\lim \frac{1}{n}\left[\underbrace{(x+x+...+x)}_{n-1}+\frac{a(1+2+...+n-1)}{n} \right]\)
\(=\lim \frac{1}{n}[(n-1)x+a(n-1)]=\lim \frac{n-1}{n}(x+a)=\lim (1-\frac{1}{n})(x+a)\)
\(=x+a\)
15 tháng 10 2023
1:
\(K=\lim\limits_{n\rightarrow\infty}\dfrac{3\cdot2^n-3^n}{2^{n+1}+3^{n+1}}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3\cdot2^n-3^n}{2^n\cdot2+3^n\cdot3}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3\cdot\dfrac{2^n}{3^n}-1}{\left(\dfrac{2}{3}\right)^n\cdot2+3}\)
\(=-\dfrac{1}{3}\)
2:
\(\lim\limits_{n\rightarrow\infty}\dfrac{3^n-4^{n+1}}{3^{n+2}+4^n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3^n-4^n\cdot4}{3^n\cdot9+4^n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{\left(\dfrac{3}{4}\right)^n-4}{\left(\dfrac{3}{4}\right)^n\cdot9+1}=-\dfrac{4}{1}=-4\)
24 tháng 11 2023
a: \(\lim\limits_{n\rightarrow+\infty}\dfrac{n^5+n^2-n+2}{\left(2n^3-1\right)\left(n^2+n+1\right)}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{1+\dfrac{1}{n^3}-\dfrac{1}{n^4}+\dfrac{2}{n^5}}{\left(\dfrac{2n^3}{n^3}-\dfrac{1}{n^3}\right)\left(\dfrac{n^2+n+1}{n^2}\right)}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{1+\dfrac{1}{n^3}-\dfrac{1}{n^4}+\dfrac{2}{n^5}}{\left(2-\dfrac{1}{n^3}\right)\left(1+\dfrac{1}{n}+\dfrac{1}{n^2}\right)}\)
\(=\dfrac{1}{2\cdot1}=\dfrac{1}{2}\)
b: \(\lim\limits_{n\rightarrow+\infty}\dfrac{\sqrt{n^2-n+2}}{n+2}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{n\sqrt{1-\dfrac{1}{n}+\dfrac{2}{n^2}}}{n\left(1+\dfrac{2}{n}\right)}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{\sqrt{1-\dfrac{1}{n}+\dfrac{2}{n^2}}}{1+\dfrac{2}{n}}=\dfrac{\sqrt{1-0+0}}{1+0}=\dfrac{1}{1}=1\)
c: \(\lim\limits_{n\rightarrow+\infty}\dfrac{n-\sqrt[3]{n^2-n^3}}{n^2+n+1}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{\dfrac{n}{n^2}-\dfrac{\sqrt[3]{n^2-n^3}}{n^2}}{1+\dfrac{1}{n}+\dfrac{1}{n^2}}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{\dfrac{1}{n}-\sqrt[3]{\dfrac{1}{n^4}-\dfrac{1}{n^3}}}{1+\dfrac{1}{n}+\dfrac{1}{n^2}}=\dfrac{0}{1}=0\)
d: \(\lim\limits_{n\rightarrow+\infty}\left(n-\sqrt{n^2+n+1}\right)\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{n^2-n^2-n-1}{n+\sqrt{n^2+n+1}}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{-n-1}{n+\sqrt{n^2+n+1}}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{-1-\dfrac{1}{n}}{1+\sqrt{1+\dfrac{1}{n}+\dfrac{1}{n^2}}}=-\dfrac{1}{1+1}=-\dfrac{1}{2}\)
AH
Akai Haruma
Giáo viên
14 tháng 5 2021
Lời giải:
a) \(\lim\limits_{x\to -\infty}\frac{x+3}{3x-1}=\lim\limits_{x\to -\infty}\frac{1+\frac{3}{x}}{3-\frac{1}{x}}=\frac{1}{3}\)
b)
\(\lim\limits_{x\to +\infty}\frac{(\sqrt{x^2+1}+x)^n-(\sqrt{x^2+1}-x)^n}{x}=\lim\limits_{x\to +\infty} 2[(\sqrt{x^2+1}+x)^{n-1}+(\sqrt{x^2+1}+x)^{n-1}(\sqrt{x^2+1}-x)+....+(\sqrt{x^2+1}-x)^{n-1}]\)
\(=+\infty\)
1. \(lim_{n\rightarrow+\infty}\dfrac{n^2+1}{n}=lim_{n\rightarrow+\infty}\left(n+\dfrac{1}{n}\right)=+\infty\)(đpcm)
2. \(lim_{n\rightarrow+\infty}\dfrac{2-n}{\sqrt{n}}=lim_{n\rightarrow+\infty}\left(2-\sqrt{n}\right)=-\infty\) (đpcm)
cảm ơn
( bài này học từ 1 năm trc )