Câu 4 ( 1,0 điểm). Tính $\lim _{x \rightarrow-\infty} \frac{4 x-5}{\sqrt{x^2+x+1}-x-1}$.

`\lim_{x\rightarrow-\infty}\frac{4x-5}{\sqrt{x^{2}+x+1}-x-1} = \lim_{x\rightarrow-\infty}\frac{(4x-5)(\sqrt{x^{2}+x+1}+x+1)}{(\sqrt{x^{2}+x+1}-x-1)(\sqrt{x^{2}+x+1}+x+1)}`

`= \lim_{x\rightarrow-\infty}\frac{(4x-5)(\sqrt{x^{2}+x+1}+x+1)}{x^{2}+x+1-(x+1)^{2}}`

`= \lim_{x\rightarrow-\infty}\frac{(4x-5)(\sqrt{x^{2}+x+1}+x+1)}{x^{2}+x+1-x^{2}-2x-1}`

`= \lim_{x\rightarrow-\infty}\frac{(4x-5)(\sqrt{x^{2}+x+1}+x+1)}{-x}`

Vì `x \rightarrow -\infty` ta có thể giả sử `x < 0`. Khi đó `\sqrt{x^2} = |x| = -x`

`\lim_{x\rightarrow-\infty}\frac{(4x-5)(\sqrt{x^{2}+x+1}+x+1)}{-x} = \lim_{x\rightarrow-\infty}\frac{(4x-5)(\sqrt{x^2(1+\frac{1}{x}+\frac{1}{x^2})}+x+1)}{-x}`

`= \lim_{x\rightarrow-\infty}\frac{(4x-5)(-x\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+x+1)}{-x}`

`= \lim_{x\rightarrow-\infty}\frac{(4x-5)(-x(1+\frac{1}{2x}+o(\frac{1}{x}))+x+1)}{-x}`

`= \lim_{x\rightarrow-\infty}\frac{(4x-5)(-x-\frac{1}{2}+x+1)}{-x}`

`= \lim_{x\rightarrow-\infty}\frac{(4x-5)(\frac{1}{2})}{-x}`

`= \lim_{x\rightarrow-\infty}\frac{2x-\frac{5}{2}}{-x}`

`= \lim_{x\rightarrow-\infty}(-2+\frac{5}{2x})`

`= -2`