Cho hàm số

A

- Hướng dẫn giải

Chọn A

$\begin{array}{l}\frac{1}{x^3-1}-\frac{1}{x-1}\\ =\frac{1}{(x-1).(x^2+x+1)}-\text{}\frac{1}{x-1}\\ =\frac{1-(x^2+\text{}x+1)}{(x-1).(x^2+x+1)}\\ =\frac{-x^2-x}{(x-1).(x^2+x+1)}\\ =\frac{-x^2-x}{x^3-1}\end{array}$

$\underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }\left(\frac{-x^2-x}{x^3-1}\right)$

$\underset{x\to 1^{+}}{\lim }\left(-x^2-x\right)=-2; \underset{x\to 1^{+}}{\lim }\left(x^3-1\right)= 0$

khi $x\to 1^{+}\Rightarrow x>1\Rightarrow x^3-1>0$

vậy $\underset{x\to 1^{+}}{\lim }f\left(x\right)=-\infty$