Question

tìm a,b sao cho

\(\lim_{x\rightarrow-2}\dfrac{ax^{3}+bx^{2}+4}{(x-1)^{2}(x+2)}=2\)

 

Answer 1

Giới hạn đã cho hữu hạn nên \(ax^3+bx^2+4=0\) có nghiệm \(x=-2\)

\(\Rightarrow-8a+4b+4=0\Rightarrow b=2a-1\)

\(\lim\limits_{x\rightarrow-2}\dfrac{ax^3+\left(2a-1\right)x^2+4}{\left(x-1\right)^2\left(x+2\right)}=\lim\limits_{x\rightarrow-2}\dfrac{\left(x+2\right)\left(ax^2-x+2\right)}{\left(x-1\right)^2\left(x+2\right)}\)

\(=\lim\limits_{x\rightarrow-2}\dfrac{ax^2-x+2}{\left(x-1\right)^2}=\dfrac{4a+4}{9}=2\Rightarrow a=\dfrac{7}{2}\) \(\Rightarrow b=6\)