Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(\lim\limits\dfrac{5n+1}{2n}=\lim\limits\dfrac{\dfrac{5n}{n}+\dfrac{1}{n}}{\dfrac{2n}{n}}=\lim\limits\dfrac{5+\dfrac{1}{n}}{2}=\dfrac{5+0}{2}=\dfrac{5}{2}\)
b: \(\lim\limits\dfrac{6n^2+8n+1}{5n^2+3}\)
\(=\lim\limits\dfrac{\dfrac{6n^2}{n^2}+\dfrac{8n}{n^2}+\dfrac{1}{n^2}}{\dfrac{5n^2}{n^2}+\dfrac{3}{n^2}}\)
\(=\lim\limits\dfrac{6+\dfrac{8}{n}+\dfrac{1}{n^2}}{5+\dfrac{3}{n^2}}\)
\(=\dfrac{6+0+0}{5+0}=\dfrac{6}{5}\)
c: \(\lim\limits\dfrac{3^n+2^n}{4\cdot3^n}\)
\(=\lim\limits\dfrac{\dfrac{3^n}{3^n}+\left(\dfrac{2}{3}\right)^n}{4\cdot\left(\dfrac{3^n}{3^n}\right)}\)
\(=\lim\limits\dfrac{1+\left(\dfrac{2}{3}\right)^n}{4}=\dfrac{1+0}{4}=\dfrac{1}{4}\)
d: \(\lim\limits\dfrac{\sqrt{n^2+5n+3}}{6n+2}\)
\(=\lim\limits\dfrac{\sqrt{\dfrac{n^2}{n^2}+\dfrac{5n}{n^2}+\dfrac{3}{n^2}}}{\dfrac{6n}{n}+\dfrac{2}{n}}\)
\(=\lim\limits\dfrac{\sqrt{1+\dfrac{5}{n}+\dfrac{3}{n^2}}}{6+\dfrac{2}{n}}\)
\(=\dfrac{\sqrt{1+0+0}}{6}=\dfrac{1}{6}\)
\(a,lim\dfrac{5n+1}{2n}=lim\dfrac{\dfrac{5n}{n}+\dfrac{1}{n}}{\dfrac{2n}{n}}=lim\dfrac{5+\dfrac{1}{n}}{2}=\dfrac{5}{2}\\ b,lim\dfrac{6n^2+8n+1}{5n^2+3}=lim\dfrac{\dfrac{6n^2}{n^2}+\dfrac{8n}{n^2}+\dfrac{1}{n^2}}{\dfrac{5n^2}{n^2}+\dfrac{3}{n^2}}=lim\dfrac{6+\dfrac{8}{n}+\dfrac{1}{n^2}}{5+\dfrac{3}{n^2}}=\dfrac{6}{5}\)
\(c,lim\dfrac{3^n+2^n}{4.3^n}=\dfrac{\dfrac{3^n}{3^n}+\dfrac{2^n}{3^n}}{\dfrac{4.3^n}{3^n}}=\dfrac{1+\left(\dfrac{2}{3}\right)^n}{4}=\dfrac{1}{4}\)
\(d,lim\dfrac{\sqrt{n^2+5n+3}}{6n+2}=lim\dfrac{\sqrt{\dfrac{n^2+5n+3}{n^2}}}{\dfrac{6n}{n}+\dfrac{2}{n}}=lim\dfrac{\sqrt{1+\dfrac{5}{n}+\dfrac{3}{n^2}}}{6+\dfrac{2}{n}}=\dfrac{1}{6}\)
\(a=lim\dfrac{\left(\dfrac{2}{6}\right)^n+1-\dfrac{1}{4}\left(\dfrac{4}{6}\right)^n}{\left(\dfrac{3}{6}\right)^n+6}=\dfrac{1}{6}\)
\(b=\lim\dfrac{\left(n+1\right)^2}{3n^2+4}=\lim\dfrac{n^2+2n+1}{3n^2+4}=\lim\dfrac{1+\dfrac{2}{n}+\dfrac{1}{n^2}}{3+\dfrac{4}{n^2}}=\dfrac{1}{3}\)
\(c=\lim\dfrac{n\left(n+1\right)}{2\left(n^2-3\right)}=\lim\dfrac{n^2+n}{2n^2-6}=\lim\dfrac{1+\dfrac{1}{n}}{2-\dfrac{6}{n^2}}=\dfrac{1}{2}\)
\(d=\lim\left[1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right]=\lim\left[1-\dfrac{1}{n+1}\right]=1\)
\(e=\lim\dfrac{1}{2}\left[1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right]\)
\(=\lim\dfrac{1}{2}\left[1-\dfrac{1}{2n+1}\right]=\dfrac{1}{2}\)
Một câu thôi: Liên hợp
\(\dfrac{1}{2\sqrt{1}+\sqrt{2}}=\dfrac{2.1-\sqrt{2}}{2^2-2}=\dfrac{2-\sqrt{2}}{2}=1-\dfrac{1}{\sqrt{2}}\)
\(\dfrac{1}{3\sqrt{2}+2\sqrt{3}}=\dfrac{3\sqrt{2}-2\sqrt{3}}{9.2-4.3}=\dfrac{3\sqrt{2}-2\sqrt{3}}{6}=\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\)
\(\Rightarrow\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Nên chứng minh bằng quy nạp mạnh cho chặt chẽ, giờ tui buồn ngủ quá nên bạn tự chứng minh nha :(
\(\Rightarrow u_n=1-\dfrac{1}{\sqrt{n+1}}=\dfrac{\sqrt{n+1}-1}{\sqrt{n+1}}\Rightarrow\lim\limits\left(u_n\right)=\lim\limits\dfrac{\sqrt{\dfrac{n}{n}+\dfrac{1}{n}}-\dfrac{1}{\sqrt{n}}}{\sqrt{\dfrac{n}{n}+\dfrac{1}{n}}}=1\)
\(\frac{n^3-1}{n^3+1}=\frac{\left(n-1\right)\left(n^2+n+1\right)}{\left(n+1\right)\left(n^2-n+1\right)}=\frac{\left(n-1\right)\left[\left(n+1\right)^2-\left(n+1\right)+1\right]}{\left(n+1\right)\left(n^2-n+1\right)}\)
\(\Rightarrow u_n=\frac{1.\left(3^2-3+1\right)}{3.\left(2^2-2+1\right)}.\frac{2\left(4^2-4+1\right)}{4.\left(3^2-3+1\right)}.\frac{3\left(5^2-5+1\right)}{5\left(4^2-4+1\right)}...\frac{\left(n-1\right)\left[\left(n+1\right)^2-\left(n+1\right)+1\right]}{\left(n+1\right)\left(n^2-n+1\right)}\)
\(\Rightarrow u_n=\frac{1.2.\left[\left(n+1\right)^2-\left(n+1\right)+1\right]}{\left(2^2-2+1\right).n\left(n+1\right)}=\frac{2n^2+2n+2}{3n^2+3n}\)
\(\Rightarrow lim\left(u_n\right)=lim\frac{2n^2+2n+2}{3n^2+3n}=\frac{2}{3}\)
a/
\(u_n=\dfrac{1}{\left(2-1\right)\left(2+1\right)}+\dfrac{1}{\left(3-1\right)\left(3+1\right)}+...+\dfrac{1}{\left(n-1\right)\left(n+1\right)}\)
\(u_n=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+\dfrac{1}{4.6}+...+\dfrac{1}{\left(n-2\right)n}+\dfrac{1}{\left(n-1\right)\left(n+1\right)}\)
\(u_n=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{n-2}-\dfrac{1}{n}+\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)\)
\(u_n=\dfrac{1}{2}\left(1+\dfrac{1}{2}-\dfrac{1}{n}-\dfrac{1}{n+1}\right)=\dfrac{1}{2}\left(\dfrac{3}{2}-\dfrac{1}{n}-\dfrac{1}{n+1}\right)\)
\(\Rightarrow lim\left(u_n\right)=lim\left(\dfrac{1}{2}\left(\dfrac{3}{2}-\dfrac{1}{n}-\dfrac{1}{n+1}\right)\right)=\dfrac{1}{2}.\dfrac{3}{2}=\dfrac{3}{4}\)
b/ \(u_n=\dfrac{1}{1^2+3}+\dfrac{1}{2^2+6}+...+\dfrac{1}{n^2+3n}=\dfrac{1}{1.4}+\dfrac{1}{2.5}+...+\dfrac{1}{n\left(n+3\right)}\)
\(u_n=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{n}-\dfrac{1}{n+3}\right)\)
\(u_n=\dfrac{1}{3}\left(1+\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{n+1}-\dfrac{1}{n+2}-\dfrac{1}{n+3}\right)\)
\(\Rightarrow lim\left(u_n\right)=lim\left(\dfrac{1}{3}\left(1+\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{n+1}-\dfrac{1}{n+2}-\dfrac{1}{n+3}\right)\right)\)
\(\Rightarrow lim\left(u_n\right)=\dfrac{1}{3}\left(1+\dfrac{1}{2}+\dfrac{1}{3}\right)=\dfrac{11}{18}\)
1/ \(\lim\limits\dfrac{\dfrac{2^n}{7^n}-5.7.\left(\dfrac{7}{7}\right)^n}{\dfrac{2^n}{7^n}+\left(\dfrac{7}{7}\right)^n}=-35\)
2/ \(\lim\limits\dfrac{\dfrac{3^n}{7^n}-2.5.\left(\dfrac{5}{7}\right)^n}{\dfrac{2^n}{7^n}+\dfrac{7^n}{7^n}}=0\)
3/ \(\lim\limits\sqrt[3]{\dfrac{\dfrac{5}{n}-\dfrac{8n}{n}}{\dfrac{n}{n}+\dfrac{3}{n}}}=\sqrt[3]{-8}=-2\)
a) \(\lim \frac{{5n + 1}}{{2n}} = \lim \frac{{5 + \frac{1}{n}}}{2} = \frac{{5 + 0}}{2} = \frac{5}{2}\)
b) \(\lim \frac{{6{n^2} + 8n + 1}}{{5{n^2} + 3}} = \lim \frac{{6 + \frac{8}{n} + \frac{1}{{{n^2}}}}}{{5 + \frac{3}{{{n^2}}}}} = \frac{{6 + 0 + 0}}{{5 + 0}} = \frac{6}{5}\)
c) \(\lim \frac{{\sqrt {{n^2} + 5n + 3} }}{{6n + 2}} = \lim \frac{{\sqrt {1 + \frac{5}{n} + \frac{3}{{{n^2}}}} }}{{6 + \frac{2}{n}}} = \frac{{\sqrt {1 + 0 + 0} }}{{6 + 0}} = \frac{1}{6}\)
d) \(\lim \left( {2 - \frac{1}{{{3^n}}}} \right) = \lim 2 - \lim {\left( {\frac{1}{3}} \right)^n} = 2 - 0 = 0\)
e) \(\lim \frac{{{3^n} + {2^n}}}{{{{4.3}^n}}} = \lim \frac{{1 + {{\left( {\frac{2}{3}} \right)}^n}}}{4} = \frac{{1 + 0}}{4} = \frac{1}{4}\)
g) \(\lim \frac{{2 + \frac{1}{n}}}{{{3^n}}}\)
Ta có \(\lim \left( {2 + \frac{1}{n}} \right) = \lim 2 + \lim \frac{1}{n} = 2 + 0 = 2 > 0;\lim {3^n} = + \infty \Rightarrow \lim \frac{{2 + \frac{1}{n}}}{{{3^n}}} = 0\)
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:
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3^n-4^n\cdot4}{3^n\cdot9+4^n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{3^n}{4^n}-4}{3^n\cdot\dfrac{9}{4^n}+1}\)
\(=-\dfrac{4}{1}=-4\)
