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

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

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

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

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

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

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

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.

\(a=\lim\dfrac{1}{\sqrt{4n+1}+2\sqrt{n}}=\dfrac{1}{\infty}=0\)

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

\(c=\lim4^n\left(\sqrt{\left(\dfrac{9}{16}\right)^n-\left(\dfrac{3}{16}\right)^n}-1\right)=+\infty.\left(-1\right)=-\infty\)

\(d=\lim n^3\left(3+\dfrac{2}{n}+\dfrac{1}{n^2}\right)=+\infty.3=+\infty\)

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 \(n^3\))

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

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

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

b) \(lim\dfrac{4^n}{2.3^n+4^n}=lim\dfrac{1}{2.\left(\dfrac{3}{4}\right)^n+1}=\dfrac{1}{1}=1\).

\(a=\lim\dfrac{17+\dfrac{3}{n}+\dfrac{4}{n^3}}{2+\dfrac{1}{n^2}}=\dfrac{17}{2}\)

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

\(=\lim\left(\dfrac{2n^3-13n^2+3}{\left(2n^2+3\right)\left(5n+1\right)}\right)=\lim\dfrac{2-\dfrac{13}{n}+\dfrac{3}{n^3}}{\left(2+\dfrac{3}{n^2}\right)\left(5+\dfrac{1}{n}\right)}=\dfrac{2}{2.5}=\dfrac{1}{5}\)

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

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

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

Lim 3.4n-2.13n/5n+6.13n

\(a,lim\dfrac{2n^2+1}{3n^3-3n+3}\)

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

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

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

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