Question

tính\(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt[3]{x}+x^2+x+1}{\sqrt[3]{8x}+\sqrt{4x^2+x+1}}\)

Answer 1

\(=\lim\limits_{x\rightarrow-1}\dfrac{\sqrt[3]{x}+1+x^2+x}{\sqrt[3]{8x}+2+\sqrt{4x^2+x+1}-2}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{x+1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}+x\left(x+1\right)}{\dfrac{8\left(x+1\right)}{\sqrt[3]{64x^2}-2\sqrt[3]{8x}+4}+\dfrac{\left(x+1\right)\left(4x-3\right)}{\sqrt[]{4x^2+x+1}+2}}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}+x}{\dfrac{8}{\sqrt[3]{64x^2}-2\sqrt[3]{8x}+4}+\dfrac{4x-3}{\sqrt[]{4x^2+x+1}+2}}\)

\(=...\) (bạn tự thay số bấm máy)