Question

giá trị thực m sao cho lim (x-> \(-\infty\)) \(\left(\dfrac{\left(2x^2-1\right)\left(mx+3\right)}{x^3+4x+7}=6\right)\)

Answer 1

\(\lim\limits_{x\rightarrow-\infty}\dfrac{\left(2x^2-1\right)\left(mx+3\right)}{x^3+4x+7}=\lim\limits_{x\rightarrow-\infty}\dfrac{\left(2-\dfrac{1}{x^2}\right)\left(m+\dfrac{3}{x}\right)}{1+\dfrac{4}{x^2}+\dfrac{7}{x^3}}=2m\)

\(\Rightarrow2m=6\Rightarrow m=3\)