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

1: \(-1< =cosx< =1\)

=>\(-3< =3\cdot cosx< =3\)

=>\(y\in\left[-3;3\right]\)

2:

TXĐ là D=R

3: \(L=\lim\limits\dfrac{-3n^3+n^2}{2n^3+5n-2}\)

\(=\lim\limits\dfrac{-3+\dfrac{1}{n}}{2+\dfrac{5}{n^2}-\dfrac{2}{n^3}}=-\dfrac{3}{2}\)

4:

\(L=lim\left(3n^2+5n-3\right)\)

\(=\lim\limits\left[n^2\left(3+\dfrac{5}{n}-\dfrac{3}{n^2}\right)\right]\)

\(=+\infty\) vì \(\left\{{}\begin{matrix}lim\left(n^2\right)=+\infty\\\lim\limits\left(3+\dfrac{5}{n}-\dfrac{3}{n^2}\right)=3>0\end{matrix}\right.\)

5:

\(\lim\limits_{n\rightarrow+\infty}n^3-2n^2+3n-4\)

\(=\lim\limits_{n\rightarrow+\infty}n^3\left(1-\dfrac{2}{n}+\dfrac{3}{n^2}-\dfrac{4}{n^3}\right)\)

\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{n\rightarrow+\infty}n^3=+\infty\\\lim\limits_{n\rightarrow+\infty}1-\dfrac{2}{n}+\dfrac{3}{n^2}-\dfrac{4}{n^3}=1>0\end{matrix}\right.\)

\(1,y=3cosx\)

\(+TXD\) \(D=R\)

Có \(-1\le cosx\le1\)

\(\Leftrightarrow-3\le3cosx\le3\)

Vậy có tập giá trị \(T=\left[-3;3\right]\)

\(2,y=cosx\)

\(TXD\) \(D=R\)

\(3,L=lim\dfrac{n^2-3n^3}{2n^3+5n-2}=lim\dfrac{\dfrac{1}{n}-3}{2+\dfrac{5}{n^2}-\dfrac{2}{n^3}}\)(chia cả tử và mẫu cho \(n^3\))

\(=\dfrac{lim\dfrac{1}{n}-lim3}{lim2+5lim\dfrac{1}{n^2}-2lim\dfrac{1}{n^3}}=\dfrac{0-3}{2+5.0-2.0}=-\dfrac{3}{2}\)

\(4,L=lim\left(3n^2+5n-3\right)\\ =lim\left(3+\dfrac{5}{n}-\dfrac{3}{n^2}\right)\\ =lim3+5lim\dfrac{1}{n}-3lim\dfrac{1}{n^2}\\ =3\)

\(5,\lim\limits_{n\rightarrow+\infty}\left(n^3-2n^2+3n-4\right)\\ =lim\left(1-\dfrac{2}{n}+\dfrac{3}{n^2}-\dfrac{4}{n^3}\right)\\ =lim1-0\\ =1\)

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

\(\lim\dfrac{n^4-3n+4}{an^3+2n^2+1}=\lim\dfrac{n-\dfrac{3}{n^2}+\dfrac{4}{n^3}}{a+\dfrac{2}{n}+\dfrac{1}{n^3}}=+\infty.\left(\dfrac{1}{a}\right)\)

Giới hạn đã cho bằng \(-\infty\) khi và chỉ khi \(\dfrac{1}{a}< 0\Leftrightarrow a< 0\)

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

\(B=\lim\limits\dfrac{2n+3}{n^2+1}=\lim\limits\dfrac{\dfrac{2n}{n^2}+\dfrac{3}{n^2}}{\dfrac{n^2}{n^2}+\dfrac{1}{n^2}}=0\)

- Ta có:

- Suy ra  C = 2 4 = 16 .

Chọn C.

Đáp án C.

- Ta có:

Chọn C.