Tính giới hạn \(\lim\limits_{x\to2}\sqrt[3]{\dfrac{x^2-x-1}{x^2+2x}}\).
\(\dfrac{1}{4}\) | |
\(\dfrac{1}{2}\) | |
\(\dfrac{1}{3}\) | |
\(\dfrac{1}{5}\) |
Chọn phương án B.
\(\lim\limits_{x\to2}\sqrt[3]{\dfrac{x^2-x-1}{x^2+2x}}=\sqrt[3]{\dfrac{2^2-2-1}{2^2+2\cdot2}}=\dfrac{1}{2}\).