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

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

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

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

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

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

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

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.

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:

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

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

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

1:

\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2-1-9n^2}{\sqrt{n^2-1}-3n}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{-8n^2-1}{\sqrt{n^2-1}-3n}\)

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

2: 

\(\lim\limits_{n\rightarrow\infty}\sqrt{4n^2+5}+n\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{4n^2+5-n^2}{\sqrt{4n^2+5}-n}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{3n^2+5}{\sqrt{4n^2+5}-n}\)

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

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

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 \(n^3\))

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

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

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

Lời giải:
1.

\(\lim\limits_{n\to \infty}(\sqrt{n^2+6n}-n)=\lim\limits_{n\to \infty}\frac{6n}{\sqrt{n^2+6n}+n}=\lim\limits_{n\to \infty}\frac{6}{\sqrt{1+\frac{6}{n}}+1}=\frac{6}{1+1}=3\)

2.

\(\lim\limits_{n\to \infty}(\sqrt{n+1}-\sqrt{n-1})=\lim\limits_{n\to \infty}\frac{(n+1)-(n-1)}{\sqrt{n+1}+\sqrt{n-1}}=\lim\limits_{n\to \infty}\frac{2}{\sqrt{n+1}+\sqrt{n-1}}=0\) do $\sqrt{n+1}+\sqrt{n-1}\to \infty$ khi $n\to \infty$

Đáp án B

lim x → − ∞ x + x 2 + x x + 2 = lim x → − ∞ x + x 1 + 1 x 2 x 1 + 2 x 2 = lim x → − ∞ − 2 x x = − 2

Đáp án B

Phương pháp : Chia cả tử và mẫu cho x và sử dụng giới hạn 

Cách giải : 

1.

Câu này chắc người ra đề nhầm lẫn, vì giới hạn đã cho không tồn tại (giới hạn phải tại 1 bằng dương vô cực, giới hạn trái tại 1 bằng âm vô cực nên ko tồn tại giới hạn tại 1)

Nếu đề là \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[]{2x-1}\sqrt[3]{3x-2}-1}{x-1}\) thì tính được

2.

\(\lim\limits_{x\rightarrow2}\dfrac{\sqrt[]{2x-3}.\sqrt[3]{5x-2}-2}{x-2}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\sqrt[3]{5x-2}\left(\sqrt[]{2x-3}-1\right)+\sqrt[3]{5x-2}-2}{x-2}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\dfrac{\sqrt[3]{5x-2}\left(2x-4\right)}{\sqrt[]{2x-3}+1}+\dfrac{5x-10}{\sqrt[3]{\left(5x-2\right)^3}+2\sqrt[3]{5x-2}+4}}{x-2}\)

\(=\lim\limits_{x\rightarrow2}\left(\dfrac{2\sqrt[3]{5x-2}}{\sqrt[]{2x-3}+1}+\dfrac{5}{\sqrt[3]{\left(5x-3\right)^2}+2\sqrt[3]{5x-3}+4}\right)\)

\(=\dfrac{2.2}{1+1}+\dfrac{5}{4+4+4}=...\)

Câu 3 đề bài phạm sai lầm y như câu 1

Lời giải:

Với $x\to -\infty$ thì $3x^2-5x=x^2(3-\frac{5}{x})\to +\infty$ do $x^2\to +\infty$ và $3-\frac{5}{x}\to 3 >0$

$\Rightarrow \lim\limits_{x\to -\infty}\sqrt{3x^2-5x}=+\infty$