Sao anh không thấy đề cụ thể ta!

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

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

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

a: \(=lim_{x->-\infty}\dfrac{2x-5+\dfrac{1}{x^2}}{7-\dfrac{1}{x}+\dfrac{4}{x^2}}\)

\(=\dfrac{2x-5}{7}\)

\(=\dfrac{2}{7}x-\dfrac{5}{7}\)

\(=-\infty\)

b: \(=lim_{x->+\infty}x\sqrt{\dfrac{1+\dfrac{1}{x}+\dfrac{3}{x^2}}{3x^2+4-\dfrac{5}{x^2}}}\)

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

a: \(=\lim\limits_{x\rightarrow+\infty}\dfrac{4+\dfrac{3}{x}}{2}=\dfrac{4}{2}=2\)

b: \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{2}{x}}{3+\dfrac{1}{x}}=0\)

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

`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`

Bài 1:

\(a=\lim\limits_{x\rightarrow-\infty}\frac{2\left|x\right|+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2x+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2+\frac{1}{x}}{3-\frac{1}{x}}=-\frac{2}{3}\)

\(b=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}}{1+\frac{1}{x}}=\frac{\sqrt{9}-\sqrt{4}}{1}=1\)

\(c=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{2}{x}+\frac{3}{x^2}}+4+\frac{1}{x}}{\sqrt{4+\frac{1}{x^2}}+\frac{2}{x}-1}=\frac{1+4}{\sqrt{4}-1}=5\)

\(d=\lim\limits_{x\rightarrow+\infty}\frac{\frac{3}{x}-\frac{2}{x\sqrt{x}}+\sqrt{1-\frac{5}{x^3}}}{2+\frac{4}{x}-\frac{5}{x^2}}=\frac{1}{2}\)

Bài 2:

\(a=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{1}{x}}{1-\frac{1}{x}}=2\)

\(b=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{3}{x^3}}{1-\frac{2}{x}+\frac{1}{x^3}}=2\)

\(c=\lim\limits_{x\rightarrow+\infty}\frac{x^2\left(3+\frac{1}{x^2}\right)x\left(5+\frac{3}{x}\right)}{x^3\left(2-\frac{1}{x^3}\right)x\left(1+\frac{4}{x}\right)}=\frac{15}{+\infty}=0\)

Lời giải:
\(\lim\limits_{x\to+\infty}\frac{x-\sqrt{4x^2+x-1}}{3x+2}=\lim\limits_{x\to+\infty}\frac{1-\sqrt{4+\frac{1}{x}-\frac{1}{x^2}}}{3+\frac{2}{x}}=\frac{1-2}{3}=\frac{-1}{3}\)

`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`

\(=\frac{\left|x\right|\sqrt{1+\frac{2}{x}}+3x}{\left|x\right|\sqrt{4+\frac{1}{x^2}}-x+3}=\frac{-x\left(\sqrt{1+\frac{2}{x}}-3\right)}{-x\left(\sqrt{4+\frac{1}{x^2}}+1+\frac{3}{x}\right)}=\frac{1-3}{2+1+0}=...\)