a.

\(A=\lim\frac{\sqrt[3]{n^6-7n^3-5n+8}}{n+12}=\lim \frac{\sqrt[3]{\frac{n^6-7n^3-5n+8}{n^3}}}{\frac{n+12}{n}}=\lim \frac{\sqrt[3]{n^3-7-\frac{5}{n^2}+\frac{8}{n^3}}}{1+\frac{12}{n}}\)

Ta thấy:

\(\lim\sqrt[3]{n^3-7-\frac{5}{n^2}+\frac{8}{n^3}}=\infty \)

\(\lim (1+\frac{12}{n})=1\)

Suy ra $A=\infty$

b.

\(B=\lim\frac{1}{\sqrt{3n+2}-\sqrt{2n+1}}=\lim \frac{1}{\frac{3n+2-(2n+1)}{\sqrt{3n+2}+\sqrt{2n+1}}}=\lim \frac{\sqrt{3n+2}+\sqrt{2n+1}}{n+1}\)

\(=\lim \frac{\sqrt{\frac{3n+2}{n}}+\sqrt{\frac{2n+1}{n}}}{\frac{n+1}{\sqrt{n}}}=\lim \frac{\sqrt{3+\frac{2}{n}}+\sqrt{2+\frac{1}{n}}}{\sqrt{n}+\frac{1}{\sqrt{n}}}\)

Ta thấy:

\(\lim( \sqrt{3+\frac{2}{n}}+\sqrt{2+\frac{1}{n}})=\sqrt{3}+\sqrt{2}>0\)

\(\lim (\sqrt{n}+\frac{1}{\sqrt{n}})=\infty\)

$\Rightarrow B=\infty$

1.

\(\lim (n^3+4n^2-1)=\infty\) khi $n\to \infty$

2. 

\(\lim\limits_{n\to -\infty} \frac{(n+1)\sqrt{n^2-n+1}}{3n^2+n}=\lim\limits_{n\to -\infty}\frac{-\frac{n+1}{n}.\sqrt{\frac{n^2-n+1}{n^2}}}{3+\frac{1}{n}}\\ =\lim\limits_{n\to -\infty}\frac{-(1+\frac{1}{n})\sqrt{1-\frac{1}{n}+\frac{1}{n^2}}}{3+\frac{1}{n}}=\frac{-1}{3}\)

\(\lim\limits_{n\to +\infty} \frac{(n+1)\sqrt{n^2-n+1}}{3n^2+n}=\lim\limits_{n\to +\infty}\frac{\frac{n+1}{n}.\sqrt{\frac{n^2-n+1}{n^2}}}{3+\frac{1}{n}}\\ =\lim\limits_{n\to +\infty}\frac{(1+\frac{1}{n})\sqrt{1-\frac{1}{n}+\frac{1}{n^2}}}{3+\frac{1}{n}}=\frac{1}{3}\)

3.

\(\lim \frac{1+2+...+n}{2n^2}=\lim \frac{n(n+1)}{4n^2}=\lim \frac{n^2+n}{4n^2}\\ =\lim (\frac{1}{4}+\frac{1}{4n})=\frac{1}{4}\)

4.

\(\lim \frac{3^n-4.2^{n-1}-10}{7.2^n+4^n}=\lim \frac{(\frac{3}{4})^n-(\frac{2}{4})^{n-1}-\frac{10}{4^n}}{7(\frac{2}{4})^n+1}\\ =\lim \frac{(\frac{3}{4})^n-(\frac{1}{2})^{n-1}-\frac{10}{4^n}}{7(\frac{1}{2})^n+1}\\ =\frac{0-0-0}{7.0+1}=0\)

`a)lim[5n^3-3n^2+1]/[1-3n^3]`

`=lim[5-3/n+1/[n^3]]/[1/[n^3]-3]`

`=5/[-3]=-5/3`

_____________________________
`b)lim[-9n+5]/[3n-3]`

`=lim[-9+5/n]/[3-3/n]`

`=[-9]/3=-3`

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 \(n^3\))

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

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

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

a) Cả tử số và mẫu số của \(\frac{7n^2-3n+12}{n^2+2n+2}\) đều dẫn đến \(\infty\) nên không thể trả lời ngay biểu thức đó  tiến đến giới hạn nào (dạng vô định \(\left(\frac{\infty}{\infty}\right)\)). Tuy nhiên sau khi chia cả tử số và mẫu số cho \(n^2\) :

\(\frac{7n^2-3n+12}{n^2+2n+2}=\frac{7-\frac{3}{n}+\frac{12}{n^2}}{1+\frac{2}{n}+\frac{2}{n^2}}\)

Ta thấy ngay tử số gần đến 7 và mẫu số gần đến 1 (vì \(\lim\limits\frac{1}{n^p}=0,p\ge1\)

Điều đó cho phép ta áp dụng công thức và thu được kết quả \(\lim\limits\frac{7n^2-3n+12}{n^2+2n+2}=\lim\limits\frac{7-\frac{3}{n}+\frac{12}{n^2}}{1+\frac{2}{n}+\frac{2}{n^2}}=7\)

b) Áp dụng công thức "Nếu tồn tại \(\lim\limits a^n,k\in\)N* thì tồn tại \(\lim\limits\left(a_n\right)^k=\left(\lim\limits a_n\right)^k\)"

ta có : 

\(\lim\limits a_n=\left[\lim\limits\left(\frac{3n^2+n-2}{4n^2+2n+7}\right)\right]^3\)

Mặt khác do \(\lim\limits\frac{3n^2+n-2}{4n^2+2n+7}=\lim\limits\frac{3+\frac{1}{n}-\frac{2}{n^2}}{4+\frac{2}{n}+\frac{7}{n^2}}=\frac{3}{4}\)

nên \(\lim\limits a_n=\left(\frac{3}{4}\right)^3=\frac{27}{64}\)

x tiến tới đâu zậy bạn?

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

\(\lim\limits\frac{1}{\sqrt{n+1}-\sqrt{n}}=\lim\limits\left(\sqrt{n+1}+\sqrt{n}\right)=+\infty\)

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

\(b=\lim\dfrac{\sqrt{\dfrac{1}{n}+\sqrt{\dfrac{1}{n^3}+\dfrac{1}{n^4}}}}{1-\dfrac{1}{\sqrt{n}}}=\dfrac{0}{1}=0\)

\(c=\lim\dfrac{\sqrt{2n^2-1+\dfrac{7}{n^2}}}{3+\dfrac{5}{n}}=\dfrac{+\infty}{3}=+\infty\)

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