Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(=lim\frac{n\sqrt{1+\frac{1}{n}-\frac{1}{n^2}}-n\sqrt{4-\frac{2}{n^2}}}{n\left(1+\frac{3}{n}\right)}=\frac{\sqrt{1+0+0}-\sqrt{4-0}}{1+0}=-1\)
\(=lim\frac{3\left(\frac{3}{7}\right)^n-\frac{1}{4}.\left(\frac{2}{7}\right)^n-5.\left(\frac{1}{7}\right)^n}{3+6.\left(\frac{1}{7}\right)^n}=\frac{3.0-\frac{1}{4}.0-5.0}{3+6.0}=0\)
\(=lim\frac{2n-4}{3n+\sqrt{9n^2-2n+4}}=lim\frac{2-\frac{4}{n}}{3+\sqrt{9-\frac{2}{n}+\frac{4}{n^2}}}=\frac{2}{3+\sqrt{9}}=\frac{1}{3}\)
Ta có công thức: \(1^2+3^2+5^2+...+\left(2n-1\right)^2=\frac{n\left(2n-1\right)\left(2n+1\right)}{3}\)
\(lim\frac{n\left(2n-1\right)\left(2n+1\right)}{3n^3}=lim\frac{1\left(2-\frac{1}{n}\right)\left(2+\frac{1}{n}\right)}{3}=\frac{1.2.2}{3}=\frac{4}{3}\)
Mặt nước trong hồ tựa như chiếc gương bầu dục phản chiếu ánh sáng trên quê hương tôi.
HOK TỐT NHA BN YÊU!
a;Chia n cả tử và mẫu
b;Chia cho n4 mà tử dần đến 0 mẫu dần đến 1 nên lim =0
lim\(\frac{3n^2+n-5}{2n^2+1}\)=lim\(\frac{n^2\left(3+\frac{1}{n}-\frac{5}{n^2}\right)}{n^2\left(2+\frac{1}{n}\right)}\)=\(\frac{3}{2}\)
lim\(\frac{\sqrt{9n^2-n}+1}{4n-2}\)=lim\(\frac{n\sqrt{9-\frac{1}{n}+\frac{1}{n^2}}}{n\left(4-\frac{2}{n}\right)}\)=lim\(\frac{\sqrt{9}}{4}\)=\(\frac{3}{2}\)
\(\lim\limits\frac{1+2^n}{2^{n+1}-16}=\lim\limits\frac{\left(\frac{1}{2}\right)^n+1}{2-16\left(\frac{1}{2}\right)^n}=\frac{0+1}{2-0}=\frac{1}{2}\)
\(\lim\limits\left(u_n\right)=\lim\limits\frac{\sqrt{16n^2-n+1}}{3n-2}=\lim\limits\frac{\sqrt{16-\frac{1}{n}+\frac{1}{n^2}}}{3-\frac{2}{n}}=\frac{\sqrt{16-0+0}}{3-0}=\frac{4}{3}\)
Xét hàm:
\(f\left(x\right)=x+x^2+x^3+...+x^n\)
Theo công thức tổng cấp số nhân ta có:
\(x+x^2+x^3+...+x^n=\dfrac{x^{n+1}-x}{x-1}\)
Đạo hàm 2 vế ta được:
\(1+2x+3x^2+...+nx^{n-1}=\dfrac{\left[\left(n+1\right)x^n-1\right]\left(x-1\right)-\left(x^{n+1}-x\right)}{\left(x-1\right)^2}\)
\(\Leftrightarrow1+2x+3x^2+...+nx^{n-1}=\dfrac{nx^{n+1}-\left(n+1\right)x^n+1}{\left(x-1\right)^2}\)
\(\Leftrightarrow x+2x^2+3x^3+...+n.x^n=\dfrac{n.x^{n+2}-\left(n+1\right)x^{n+1}+x}{\left(x-1\right)^2}\)
Thay \(x=\dfrac{1}{2}\) vào ta được:
\(S=\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{n}{2^n}=\dfrac{n.\left(\dfrac{1}{2}\right)^{n+2}-\left(n+1\right).\left(\dfrac{1}{2}\right)^{n+1}+\dfrac{1}{2}}{\left(\dfrac{1}{2}-1\right)^2}\)
\(\Rightarrow limS=lim\dfrac{n\left(\dfrac{1}{2}\right)^{n+2}-\left(n+1\right)\left(\dfrac{1}{2}\right)^{n+1}+\dfrac{1}{2}}{\left(\dfrac{1}{2}-1\right)^2}=\dfrac{0-0+\dfrac{1}{2}}{\dfrac{1}{4}}=2\)
\(lim\left(n-3-\sqrt{n^2-\sqrt{5}n+1}\right)=lim\dfrac{-6n+n\sqrt{5}+8}{n+3+\sqrt{n^2-\sqrt{5}n+1}}\)
=\(lim\dfrac{n\left(-6+\sqrt{5}+\dfrac{8}{n}\right)}{n\left(1+\dfrac{3}{n}+\sqrt{1-\dfrac{\sqrt{5}}{n}+\dfrac{1}{n^2}}\right)}=lim\dfrac{-6+\sqrt{5}+\dfrac{8}{n}}{1+\dfrac{3}{n}+\sqrt{1-\dfrac{\sqrt{5}}{n}+\dfrac{1}{n^2}}}=\dfrac{\sqrt{5}}{2}-3\)
\(\Rightarrow a=\dfrac{1}{2};b=-3\)\(\Rightarrow a+b=\dfrac{-5}{2}\)
Em học lớp 8 thôi :)) Cái này em k chắc lắm ạ, có gì sai anh chỉ nhé !
Gợi ý :
3) \(n^3+11n=n\cdot\left(n^2+11\right)=n\cdot\left(n^2-1+12\right)\)
\(=n\left(n-1\right)\left(n+1\right)+12n⋮6\)
1) \(Có:2^n-2n-1=2\left(2^{n-1}-1\right)-1>0\forall n\ge3\)
nên : \(2^n>2n+1\)
\(\lim\limits\frac{3^n+4^n+3}{4^n+2^n-1}=\lim\limits\frac{\left(\frac{3}{4}\right)^n+1+3\left(\frac{1}{4}\right)^n}{1+\left(\frac{2}{4}\right)^n-\left(\frac{1}{4}\right)^n}=\frac{0+1+0}{1+0+0}=1\)
\(\lim\limits\frac{5.2^n+9.3^n}{2.2^n+3.3^n}=\lim\limits\frac{5\left(\frac{2}{3}\right)^n+9}{2.\left(\frac{2}{3}\right)^n+3}=\frac{0+9}{0+3}=3\)
\(\lim\limits\frac{4^n-7^n}{2^n+15^n}=\lim\limits\frac{\left(\frac{4}{15}\right)^n-\left(\frac{7}{15}\right)^n}{\left(\frac{2}{15}\right)^n+1}=\frac{0-0}{0+1}=0\)
\(\lim\limits\frac{6.5^n+9^n}{3.12^n+7^n}=\lim\limits\frac{6\left(\frac{5}{12}\right)^n+\left(\frac{9}{12}\right)^n}{3+\left(\frac{7}{12}\right)^n}=\frac{0+0}{3+0}=0\)
\(\lim\limits\frac{\sqrt{5}^n}{3^n+1}=\lim\limits\frac{\left(\frac{\sqrt{5}}{3}\right)^n}{1+\left(\frac{1}{3}\right)^n}=\frac{0}{1+0}=0\)
\(\lim\limits\frac{5.5^n-3.7^n}{3.10^n+36.6^n}=\lim\limits\frac{5.\left(\frac{5}{10}\right)^n-3\left(\frac{7}{10}\right)^n}{3+36\left(\frac{6}{10}\right)^n}=\frac{0-0}{3+0}=0\)