\(0\le sin^2n\le1\) ; \(\forall n\Rightarrow\dfrac{0}{n+2}\le\dfrac{sin^2n}{n+2}\le\dfrac{1}{n+2}\)

Mà \(\lim\left(\dfrac{0}{n+2}\right)=\lim\left(\dfrac{1}{n+2}\right)=0\)

\(\Rightarrow\lim\left(\dfrac{sin^2n}{n+2}\right)=0\)

Đáp án D

Cai bai ben duoi bai nay y. Doc hieu chet lien. Ban nen xai go cong thuc de toi uu hon

\(C=\lim\limits\dfrac{n^3+1}{n\left(2n+1\right)^2}=\lim\limits\dfrac{n^3+1}{n\left(4n^2+4n+1\right)}=\lim\limits\dfrac{n^3+1}{4n^3+4n^2+n}=\lim\limits\dfrac{\dfrac{n^3}{n^3}+\dfrac{1}{n^3}}{\dfrac{4n^3}{n^3}+\dfrac{4n^2}{n^3}+\dfrac{n}{n^3}}=\dfrac{1}{4}\)

Xai cai nay go cong thuc di ban :v Doc ko hieu

\(A=\lim\limits\left(\sqrt{n^2+2n+2}+n\right)=\lim\limits\dfrac{n^2+2n+2-n^2}{\sqrt{n^2+2n+2}-n}=\dfrac{\dfrac{2n}{n}+\dfrac{2}{n}}{\sqrt{\dfrac{n^2}{n^2}+\dfrac{2n}{n^2}+\dfrac{2}{n^2}}-\dfrac{n}{n}}=\dfrac{2}{1-1}=+\infty\)

\(C=\lim\limits\dfrac{\sqrt{n^2+1}}{n+1}?\)

\(C=\lim\limits\dfrac{\sqrt{\dfrac{n^2}{n^2}+\dfrac{1}{n^2}}}{\dfrac{n}{n}+\dfrac{1}{n}}=1\)

Chọn C.

Ta có:

\(E=\lim\limits\dfrac{\sqrt{n^3+2n}+1}{n+2}=\lim\limits\dfrac{\dfrac{\left(n^3+2n\right)^{\dfrac{1}{2}}}{n}+\dfrac{1}{n}}{\dfrac{n}{n}+\dfrac{2}{n}}=\dfrac{\dfrac{n^{\dfrac{3}{2}}}{n}}{\dfrac{n}{n}}=0\)

Chọn C.

Ta có: . Mà 

Suy ra .

\(=\lim\left(\dfrac{1+2+...+n-1}{n^2}\right)=\lim\dfrac{n\left(n-1\right)}{2n^2}=\dfrac{1}{2}\)

\(F=\lim\limits\dfrac{\sqrt[4]{n^4-2n+1}+2n}{\sqrt[3]{3n^3+n}-n}=\lim\limits\dfrac{\sqrt[4]{\dfrac{n^4}{n^4}-\dfrac{2n}{n^4}+\dfrac{1}{n^4}}+\dfrac{2n}{n}}{\sqrt[3]{\dfrac{3n^3}{n^3}+\dfrac{n}{n^3}}-\dfrac{n}{n}}=\dfrac{1+2}{3-1}=\dfrac{3}{2}\)