Question

tìm giới hạn \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4x^2+2x-1}-x}{3x-2}\)

Answer 1

Sao anh không thấy đề cụ thể ta!

Answer 2

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

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

Answer 3

\(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4x^2+2x-1}-x}{3x-2}=\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4+\dfrac{2}{x}-\dfrac{1}{x^2}}+1}{-3+\dfrac{2}{x}}=\dfrac{\sqrt{4}+1}{-3}=-1\).

Answer 4

tính \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{x+1}-\sqrt{1-x}}{x}\)