Question

Cho biết \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4x^2-7x+12}}{a\left|x\right|-17}=\dfrac{2}{3}\). Tìm giá trị của \(a\).

Answer 1

\(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4x^2-7x+12}}{a\left|x\right|-17}=\lim\limits_{x\rightarrow-\infty}\dfrac{\left|x\right|\sqrt{4-\dfrac{7}{x}+\dfrac{12}{x^2}}}{\left|x\right|\left(a-\dfrac{17}{\left|x\right|}\right)}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4-\dfrac{7}{x}+\dfrac{12}{x^2}}}{a-\dfrac{17}{\left|x\right|}}=\dfrac{2}{a}\)

\(\Rightarrow\dfrac{2}{a}=\dfrac{2}{3}\Rightarrow a=3\)