Question

Để \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4x^2+x+1}+4}{mx-2}=\dfrac{1}{2}\) thì giá trị \(m\) thuộc tập hợp nào?

Answer 1

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

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

\(\Rightarrow\dfrac{-2}{m}=\dfrac{1}{2}\Rightarrow m=-4\)