Đáp án A

\(lim\left(u_n\right)=lim\frac{2a-3an}{n+2}=lim\frac{\frac{2a}{n}-3a}{1+\frac{2}{n}}=-3a\)

Để \(lim\left(u_n\right)=-\frac{1}{3}\Rightarrow-3a=-\frac{1}{3}\Rightarrow a=\frac{1}{9}\)

Chụp ảnh hoặc sử dụng gõ công thức nhé bạn. Để vầy khó hiểu lắm

1/...

2/ \(=\lim\dfrac{\dfrac{1}{n\sqrt{n}}-1}{4+\dfrac{1}{n^2\sqrt{n}}}=\dfrac{0-1}{4+0}=-\dfrac{1}{4}\) (chia cả tử-mẫu cho \(n^3\))

3/ \(=\lim\dfrac{3-\left(\dfrac{1}{4}\right)^n}{2.\left(\dfrac{3}{4}\right)^n+4\left(\dfrac{1}{4}\right)^n}=\dfrac{3-0}{2.0+3.0}=\dfrac{3}{0}=+\infty\) (chia tử mẫu cho \(4^n\))

4/ \(=\lim\dfrac{2.2^n+\dfrac{4}{3}.3^n}{1-\dfrac{1}{2}.2^n+3.3^n}=\lim\dfrac{2.\left(\dfrac{2}{3}\right)^n+\dfrac{4}{3}}{\left(\dfrac{1}{3}\right)^n-\dfrac{1}{2}.\left(\dfrac{2}{3}\right)^n+3}=\dfrac{2.0+\dfrac{4}{3}}{0-\dfrac{1}{2}.0+3}=\dfrac{4}{9}\) (chia tử mẫu  cho \(3^n\))

AM/MB=AN/NC

=>3/NC=2/3

=>NC=4,5cm

tham khảo:

\(a) 2+5+8+...+(3n−1)=n(3n+1)2 (1) Đặt Sn=2+5+8+...+(3n−1) Với n=1 ta có: S1=2=1(3.1+1)2 Giả sử (1) đúng với n=k(k≥1), tức là Sk=2+5+8+...+(3k−1)=k(3k+1)2 Ta chứng minh (1) đúng với n=k+1 hay Sk+1=(k+1)(3k+4)2 Thật vậy ta có: Sk+1=2+5+8+...+(3k−1)+[3(k+1)−1]=Sk+3k+2=k(3k+1)2+3k+2=3k2+k+6k+42=3k2+7k+42=(k+1)(3k+4)2 Vậy (1) đúng với mọi k≥1 hay (1) đúng với mọi n∈N∗ b) 3+9+27+...+3n=12(3n+1−3) (2) Đặt Sn=3+9+27+...+3n=12(3n+1−3) Với n=1, ta có: S1=3=12(32−3) (hệ thức đúng) Giả sử (2) đúng với n=k(k≥1) tức là Sk=3+9+27+...+3k=12(3k+1−3) Ta chứng minh (2) đúng với n=k+1, tức là chứng minh Sk+1=12(3k+2−3) Thật vậy, ta có: Sk+1=3+9+27+...+3k+1=Sk+3k+1=12(3k+1−3)+3k+1=32.3k+1−32=12(3k+2−3)(đpcm) Vậy (2) đúng với mọi k≥1 hay đúng với mọi n∈N∗\)

a/ \(=\lim\limits\dfrac{\sqrt{\dfrac{n}{n}+\dfrac{1}{n}}}{\dfrac{1}{\sqrt{n}}+\sqrt{\dfrac{n}{n}}}=1\)

b/ \(1+2+...+n=\dfrac{n\left(n+1\right)}{2}\)

\(\Rightarrow\lim\limits\dfrac{n\left(n+1\right)}{2n^2+4}=\lim\limits\dfrac{\dfrac{n^2}{n^2}+\dfrac{n}{n^2}}{\dfrac{2n^2}{n^2}+\dfrac{4}{n^2}}=\dfrac{1}{2}\)

c/ \(=\lim\limits\dfrac{n^2+n+1-n^2}{\sqrt{n^2+n+1}+n}=\lim\limits\dfrac{n+1}{\sqrt{n^2+n+1}+n}=\lim\limits\dfrac{\dfrac{n}{n}+\dfrac{1}{n}}{\sqrt{\dfrac{n^2}{n^2}+\dfrac{n}{n^2}+\dfrac{1}{n^2}}+\dfrac{n}{n}}=\dfrac{1}{1+1}=\dfrac{1}{2}\)

d/ \(=\lim\limits\left[\sqrt{n}\left(\sqrt{3-\dfrac{1}{\sqrt{n}}}-\sqrt{2-\dfrac{1}{\sqrt{n}}}\right)\right]=\lim\limits\left[\sqrt{n}\left(\sqrt{3}-\sqrt{2}\right)\right]=+\infty\)

e/ \(=\lim\limits\dfrac{n^3+2n^2-n-n^3}{\left(\sqrt[3]{n^3+2n^2}\right)^2+n.\sqrt[3]{n^3+2n^2}+n^2}=\lim\limits\dfrac{2n^2-n}{\left(n^3+2n^2\right)^{\dfrac{2}{3}}+n.\left(n^3+2n^2\right)^{\dfrac{1}{3}}+n^2}\)

\(=\dfrac{2}{1+1+1}=\dfrac{2}{3}\)

g/ \(=\lim\limits\dfrac{2^n+9.3^n}{4.3^n+8.2^n}=\lim\limits\dfrac{\left(\dfrac{2}{3}\right)^n+9.\left(\dfrac{3}{3}\right)^n}{4.\left(\dfrac{3}{3}\right)^n+8.\left(\dfrac{2}{3}\right)^n}=\dfrac{9}{4}\)