Question

Biết \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{x}-1}{x^2+ax+2}=b\) . tìm a, b viết a, b là sô thuc khác 0

Answer 1

e sửa chut ạ;  \(\lim\limits_{x\rightarrow1}\) 

Answer 2

\(b\) hữu hạn nên \(x^2+ax+2=0\) có nghiệm \(x=1\)

\(\Rightarrow1+a+2=0\Rightarrow a=-3\)

\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{x}-1}{x^2-3x+2}=\lim\limits_{x\rightarrow1}\dfrac{x-1}{\left(x-1\right)\left(x-2\right)\left(\sqrt{x}+1\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{1}{\left(x-2\right)\left(\sqrt{x}+1\right)}=-\dfrac{1}{2}\Rightarrow b=-\dfrac{1}{2}\)