Đáp án:
\(\lim\limits_{x\to 1}\dfrac{(2x-1) - \sqrt[3]{6x -5}}{x^3 - x^2 - x + 1}=2\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad \lim\limits_{x\to 1}\dfrac{(2x-1) - \sqrt[3]{6x -5}}{x^3 - x^2 - x + 1}\\
= \lim\limits_{x\to 1}\dfrac{\left[(2x-1) - \sqrt[3]{6x -5}\right]\left[(2x-1)^2 + (2x-1)\sqrt[3]{6x-5} + \sqrt[3]{(6x-5)^2}\right]}{\left(x^3 - x^2 - x + 1\right)\left[(2x-1)^2 + (2x-1)\sqrt[3]{6x-5} + \sqrt[3]{(6x-5)^2}\right]}\\
= \lim\limits_{x\to 1}\dfrac{8x^3 - 12x^2 + 4}{\left(x^3 - x^2 - x + 1\right)\left[(2x-1)^2 + (2x-1)\sqrt[3]{6x-5} + \sqrt[3]{(6x-5)^2}\right]}\\
= \lim\limits_{x\to 1}\dfrac{4(x-1)^2(2x+1)}{(x-1)^2(x+1)\left[(2x-1)^2 + (2x-1)\sqrt[3]{6x-5} + \sqrt[3]{(6x-5)^2}\right]}\\
= \lim\limits_{x\to 1}\dfrac{4(2x+1)}{(x+1)\left[(2x-1)^2 + (2x-1)\sqrt[3]{6x-5} + \sqrt[3]{(6x-5)^2}\right]}\\
= \dfrac{4(2.1+1)}{(1+1)\left[(2.1-1)^2 + (2.1-1)\sqrt[3]{6.1-5} + \sqrt[3]{(6.1-5)^2}\right]}\\
=2
\end{array}\)
