\(\lim\limits_{x\rightarrow-4}\dfrac{x^2+3x-4}{x^2+4x}\) bằng
\(\dfrac{5}{4}\).\(-\dfrac{5}{4}\).\(-\dfrac{1}{2}\).\(\dfrac{1}{2}\).Hướng dẫn giải:\(\lim\limits_{x\rightarrow-4}\dfrac{x^2+3x-4}{x^2+4x}=\lim\limits_{x\rightarrow-4}\dfrac{\left(x-1\right)\left(x+4\right)}{x\left(x+4\right)}\) \(=\lim\limits_{x\rightarrow-4}\dfrac{x-1}{x}=\dfrac{5}{4}\)
