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

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

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

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

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

a: \(\lim\limits_{x\rightarrow+\infty}\dfrac{x-2}{3-\sqrt{x^2+7}}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{1-\dfrac{2}{x}}{\dfrac{3}{x}-\sqrt{1+\dfrac{7}{x^2}}}\)

\(=\dfrac{1}{0-\sqrt{1+0}}=\dfrac{1}{-1}=-1\)

b: \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{x^2-x}-\sqrt{4x^2+1}}{2x+3}\)

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

\(=\dfrac{-x\cdot\sqrt{1-\dfrac{1}{x}}+x\cdot\sqrt{4+\dfrac{1}{x^2}}}{x\left(2+\dfrac{3}{x}\right)}\)

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

Da nan roi mang meo lam mat het bai -.-

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

2/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{8x^7}{\left(-2x^7\right)}=-\dfrac{8}{2^7}\)

3/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\left(4x^2-3x+4-4x^2\right)\left(\sqrt{x^2+x+1}+x\right)}{\left(x^2+x+1-x^2\right)\left(\sqrt{4x^2-3x+4}+2x\right)}=\dfrac{-3.2}{2}=-3\)

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_{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`

`a)lim_{x->+oo}[x+1]/[x^2+x+1]`

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

`=0`

`b)lim_{x->+oo}[3x+1]/[3x^2-x+5]`

`=lim_{x->+oo}[3/x+1/[x^2]]/[3-1/x+5/[x^2]]`

`=0`

`c)lim_{x->-oo}[3x+5]/[\sqrt{x^2+x}]`

`=lim_{x->-oo}[3+5/x]/[-\sqrt{1+1/x}]`

`=-3`

`d)lim_{x->+oo}[-5x+1]/[\sqrt{3x^2+1}]`

`=lim_{x->+oo}[-5+1/x]/[\sqrt{3+1/[x^2]}]`

`=-5/3`

Lời giải:

a) \(\lim\limits_{x\to -\infty}\frac{x+3}{3x-1}=\lim\limits_{x\to -\infty}\frac{1+\frac{3}{x}}{3-\frac{1}{x}}=\frac{1}{3}\)

b) 

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

\(=+\infty\)