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 \(n^3\))

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

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

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

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:

\(=\lim\limits_{n\rightarrow\infty}\dfrac{3^n-4^n\cdot4}{3^n\cdot9+4^n}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{3^n}{4^n}-4}{3^n\cdot\dfrac{9}{4^n}+1}\)

\(=-\dfrac{4}{1}=-4\)

Lời giải:
\(\lim\frac{6n^3-2n+1}{(5n^3-n)(n^2+n-1)}=\lim \frac{6-\frac{2}{n^2}+\frac{1}{n^3}}{(5-\frac{1}{n^2})(n^2+n-1)}\)

Ta thấy:

 \(\lim\frac{6-\frac{2}{n^2}+\frac{1}{n^3}}{5-\frac{1}{n^2}}=\frac{6}{5}\)

\(\lim \frac{1}{n^2+n-1}=0\)

$\Rightarrow L=0$

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

\(\dfrac{-n^2+2n+1}{\sqrt{3n^4+2}}=\dfrac{-\dfrac{n^2}{n^4}+\dfrac{2n}{n^4}+\dfrac{1}{n^4}}{\sqrt{\dfrac{3n^4}{n^4}+\dfrac{2}{n^4}}}=\dfrac{-\dfrac{1}{n^2}+\dfrac{2}{n^3}+\dfrac{1}{n^4}}{\sqrt{3+\dfrac{2}{n^4}}}\)

\(\Rightarrow\left(1\right)=\dfrac{-lim\dfrac{1}{n^2}+2lim\dfrac{1}{n^3}+lim\dfrac{1}{n^4}}{\sqrt{lim\left(3+\dfrac{2}{n^4}\right)}}\)

\(=\dfrac{0}{\sqrt{lim\left(3+\dfrac{2}{n^4}\right)}}=0\)

\(2,\lim\limits_{n\rightarrow\infty}\left(\dfrac{4n-\sqrt{16n^2+1}}{n+1}\right)\left(2\right)\)

\(\dfrac{4n-\sqrt{16n^2+1}}{n+1}=\dfrac{\dfrac{4n}{n^2}-\sqrt{\dfrac{16n^2}{n^2}+\dfrac{1}{n^2}}}{\dfrac{n}{n^2}+\dfrac{1}{n^2}}=\dfrac{\dfrac{4}{n}-\sqrt{16+\dfrac{1}{n^2}}}{\dfrac{1}{n}+\dfrac{1}{n^2}}\)

\(\Rightarrow\left(2\right)=\dfrac{lim\left(\dfrac{4}{n}-\sqrt{16+\dfrac{1}{n^2}}\right)}{lim\left(\dfrac{1}{n}+\dfrac{1}{n^2}\right)}=\dfrac{lim\left(\dfrac{4}{n}-\sqrt{16+\dfrac{1}{n^2}}\right)}{0}\)

Vậy giới hạn \(\left(2\right)\) không xác định.

\(3,\lim\limits_{n\rightarrow\infty}\left(\dfrac{\sqrt{9n^2+n+1}-3n}{2n}\right)\left(3\right)\)

\(\dfrac{\sqrt{9n^2+n+1}-3n}{2n}=\dfrac{\sqrt{9+\dfrac{1}{n}+\dfrac{1}{n^2}}-\dfrac{3}{n}}{\dfrac{2}{n}}\)

\(\Rightarrow\left(3\right)=\dfrac{lim\left(\sqrt{9+\dfrac{1}{n}+\dfrac{1}{n^2}}-\dfrac{3}{n}\right)}{2lim\dfrac{1}{n}}=\dfrac{lim\left(\sqrt{9+\dfrac{1}{n}+\dfrac{1}{n^2}}-\dfrac{3}{n}\right)}{0}\)

Vậy \(lim\left(3\right)\) không xác định.

2:

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

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

1:

\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^3+3n^2+1-n^3}{\sqrt[3]{n^3+3n^2+1}+n}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{3n^2+1}{\sqrt[3]{n^3+3n^2+1}+n}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(3+\dfrac{1}{n^2}\right)}{n\left(\sqrt[3]{1+\dfrac{3}{n}+\dfrac{1}{n^3}}+1\right)}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{n\cdot\left(3+\dfrac{1}{n^2}\right)}{\sqrt[3]{1+\dfrac{3}{n}+\dfrac{1}{n^3}}+1}\)

\(=\lim\limits_{n\rightarrow\infty}n\cdot\lim\limits_{n\rightarrow\infty}\dfrac{3+\dfrac{1}{n^2}}{\sqrt[3]{1+\dfrac{3}{n}+\dfrac{1}{n^3}}+1}\)

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

2: 

\(=\lim\limits_{n\rightarrow\infty}\left(\sqrt{4n^2+1}-2n+2n-\sqrt[3]{8n^3+n}\right)\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{4n^2+1-4n^2}{\sqrt{4n^2+1}+2n}+\lim\limits_{n\rightarrow\infty}\dfrac{8n^3-8n^3-n}{4n^2+2n\cdot\sqrt[3]{8n^3+n}+\left(\sqrt[3]{8n^3+n}\right)^2}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{1}{\sqrt{4n^2+1}+2n}+\lim\limits_{n\rightarrow\infty}\dfrac{-n}{4n^2+2n\cdot n\cdot\sqrt[3]{8+\dfrac{1}{n^3}}+\left(n\cdot\sqrt[3]{8+\dfrac{1}{n^3}}\right)^2}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{-n}{4n^2+2n^2\cdot\sqrt[3]{8+\dfrac{1}{n^3}}+n^2\cdot\sqrt[3]{\left(8+\dfrac{1}{n^3}\right)^2}}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{-1}{4n+2n\cdot\sqrt[3]{8+\dfrac{1}{n^3}}+n\cdot\sqrt[3]{\left(8+\dfrac{1}{n^3}\right)^2}}\)

\(=0\)

1:

\(K=\lim\limits_{n\rightarrow\infty}\dfrac{3\cdot2^n-3^n}{2^{n+1}+3^{n+1}}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{3\cdot2^n-3^n}{2^n\cdot2+3^n\cdot3}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{3\cdot\dfrac{2^n}{3^n}-1}{\left(\dfrac{2}{3}\right)^n\cdot2+3}\)

\(=-\dfrac{1}{3}\)

2: 

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

\(=\lim\limits_{n\rightarrow\infty}\dfrac{3^n-4^n\cdot4}{3^n\cdot9+4^n}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{\left(\dfrac{3}{4}\right)^n-4}{\left(\dfrac{3}{4}\right)^n\cdot9+1}=-\dfrac{4}{1}=-4\)

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

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

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

b) \(\lim \frac{{\sqrt {16{n^2} - 2} }}{n} = \lim \frac{{\sqrt {{n^2}\left( {16 - \frac{2}{{{n^2}}}} \right)} }}{n} = \lim \frac{{n\sqrt {16 - \frac{2}{{{n^2}}}} }}{n} = \lim \sqrt {16 - \frac{2}{{{n^2}}}}  = 4\)

c) \(\lim \frac{4}{{2n + 1}} = \lim \frac{4}{{n\left( {2 + \frac{1}{n}} \right)}} = \lim \left( {\frac{4}{n}.\frac{1}{{2 + \frac{1}{n}}}} \right) = \lim \frac{4}{n}.\lim \frac{1}{{2 + \frac{1}{n}}} = 0\)

d) \(\lim \frac{{{n^2} - 2n + 3}}{{2{n^2}}} = \lim \frac{{{n^2}\left( {1 - \frac{2}{n} + \frac{3}{{{n^2}}}} \right)}}{{2{n^2}}} = \lim \frac{{1 - \frac{2}{n} + \frac{3}{{{n^2}}}}}{2} = \frac{1}{2}\)