a: \(\lim_{x\to+\infty}\frac{\sqrt{x^6+2}}{3x^3-1}=\lim_{x\to+\infty}\frac{x^3\left(\sqrt{1+\frac{2}{x^6}}\right)}{x^3\left(3-\frac{1}{x^3}\right)}\)

\(=\lim_{x\to+\infty}\frac{\sqrt{1+\frac{2}{x^6}}}{3-\frac{1}{x^3}}=\frac13\)

c: \(2x^3-5x-4=2\cdot1^3-5\cdot1-4=2-5-4=2-9=-7<0\)

\(\left(x+1\right)^2=\left(1+1\right)^2=2^2=4\)

\(\lim_{x\to1}\frac{2x^3-5x-4}{\left(x+1\right)^2}=\frac{-7}{4}\)

a: \(\lim\limits_{x->0^-^-}\dfrac{-2x+x}{x\left(x-1\right)}=lim_{x->0^-}\left(\dfrac{-x}{x\left(x-1\right)}\right)\)

\(=lim_{x->0^-}\left(\dfrac{-1}{x-1}\right)=\dfrac{-1}{0-1}=\dfrac{-1}{-1}=1\)

b: \(=lim_{x->-\infty}\left(\dfrac{x^2-x-x^2+1}{\sqrt{x^2-x}+\sqrt{x^2-1}}\right)\)

\(=lim_{x->-\infty}\left(\dfrac{-x+1}{\sqrt{x^2-x}+\sqrt{x^2-1}}\right)\)

\(=lim_{x->-\infty}\left(\dfrac{-1+\dfrac{1}{x}}{-\sqrt{1-\dfrac{1}{x^2}}-\sqrt{1-\dfrac{1}{x^2}}}\right)=\dfrac{-1}{-2}=\dfrac{1}{2}\)

\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt[n]{\left(x+a_1\right)\left(x+a_2\right)...\left(x+a_n\right)}-x\right)\\ =\lim\limits_{x\rightarrow+\infty}\left(\dfrac{\left(x+a_1\right)\left(x+a_2\right)...\left(x+a_n\right)-x^n}{\sqrt[n]{\left(\left(x+a_1\right)\left(x+a_2\right)...\left(x+a_n\right)\right)^{n-1}}+...+x^{n-1}}\right)\)

= hệ số x^n-1 trên tử/hệ số x^n-1 dưới mẫu  = \(\dfrac{a_1+a_2+...+a_n}{n}\)

a: \(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x^2+x+2}}{x-1}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{1+\dfrac{1}{x}+\dfrac{2}{x^2}}}{1-\dfrac{1}{x}}=\dfrac{\sqrt{1+0+0}}{1-0}\)

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

=1

b: \(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{4x^2-x}+2x\right)\)

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

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-x}{\sqrt{x^2\left(4-\dfrac{1}{x}\right)}-2x}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-x}{-x\sqrt{4-\dfrac{1}{x}}-2x}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{1}{\sqrt{4-\dfrac{1}{x}}+2}=\dfrac{1}{\sqrt{4}+2}=\dfrac{1}{2+2}=\dfrac{1}{4}\)

Mình ko thấy đề bài

1/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2-x^2-x-x}{x+\sqrt{x^2+x+1}}=\dfrac{-2}{1-1}=-\infty\)

2/ tien toi +- vo cung?

3/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{8x^3+2x-8x^3}{\sqrt[3]{\left(8x^3+2x\right)^2}+2x.\sqrt[3]{8x^3+2x}+4x^2}=\dfrac{\dfrac{2x}{x^2}}{\dfrac{4x^2}{x^2}+\dfrac{4x^2}{x^2}+\dfrac{4x^2}{x^2}}=0\)

4/ \(\lim\limits_{x\rightarrow+\infty}\dfrac{16x^4+3x+1-16x^4}{\sqrt[4]{\left(16x^4+3x+1\right)^3}+2x.\sqrt[4]{\left(16x^4+3x+1\right)^2}+4x^2.\sqrt[4]{16x^4+3x+1}+8x^3}+\lim\limits_{x\rightarrow+\infty}\dfrac{4x^2-4x^2-2}{2x+\sqrt{4x^2+2}}=\dfrac{\dfrac{3x}{x^3}}{8+8+8+8}-\dfrac{\dfrac{2}{x}}{2+2}=0\)

5/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2+1-x^2}{\sqrt{x^2+1}+x}+\lim\limits_{x\rightarrow+\infty}\dfrac{x^2-x-x^2}{\sqrt{x^2-x}+x}=\dfrac{\dfrac{1}{x}}{1+1}-\dfrac{\dfrac{x}{x}}{1+1}=-\dfrac{1}{2}\)

Hic nan qua :( Lam vay

P/s: Anh Lam check all ho em nhung bai em lam nhe :( Em cam on

1/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2-x+1-x^2}{\sqrt{x^2-x+1}+x}=\dfrac{-1}{1+1}=-\dfrac{1}{2}\)

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

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

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

a: \(\lim\limits_{x\rightarrow+\infty}\left[x\left(\sqrt{x^2+2}-x\right)\right]\)
\(=\lim\limits_{x\rightarrow+\infty}\left[x\cdot\dfrac{x^2+2-x^2}{\sqrt{x^2+2}+x}\right]\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{2x}{\sqrt{x^2+2}+x}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{2}{\sqrt{1+\dfrac{2}{x^2}}+1}=\dfrac{2}{1+1}=\dfrac{2}{2}=1\)

b: \(\lim\limits_{x\rightarrow-\infty}\dfrac{3x^2-4x+6}{x-2}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2\left(3-\dfrac{4}{x}+\dfrac{6}{x^2}\right)}{x\left(1-\dfrac{2}{x}\right)}\)

\(=\lim\limits_{x\rightarrow-\infty}\left[x\cdot\dfrac{3-\dfrac{4}{x}+\dfrac{6}{x^2}}{1-\dfrac{2}{x}}\right]\)

\(=-\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow-\infty}x=-\infty\\\lim\limits_{x\rightarrow-\infty}\dfrac{3-\dfrac{4}{x}+\dfrac{6}{x^2}}{1-\dfrac{2}{x}}=\dfrac{3-0+0}{1-0}=\dfrac{3}{1}=3>0\end{matrix}\right.\)

`a)lim_{x->+oo} (2x-\sqrt{x^2+4x-3})`       `ĐK: x < -2-\sqrt{7};x > -2+\sqrt{7}`

`=lim_{x->+oo} [x(2-\sqrt{1+4/x -3/[x^2]}]`

`=+oo`

`b)lim_{x->+oo} (\sqrt{4x^2-3x+1}-2x)`            

`=lim_{x->+oo} [4x^2-3x+1-4x^2]/[\sqrt{4x^2-3x+1}+2x]`

`=lim_{x->+oo} [-3x+1]/[\sqrt{4x^2-3x+1}+2x]`

`=lim_{x->+oo} [-3+1/x]/[\sqrt{4-3/x+1/[x^2]}+2]`

`=-3/4`