Do mẫu số có nghiệm kép \(x=1\) và giới hạn hữu hạn \(\Rightarrow ax+b-2\sqrt{x}=0\) có nghiệm kép \(x=1\)
\(\Rightarrow a+b-2=0\Rightarrow b=2-a\)
\(\Rightarrow ax+2-a-2\sqrt{x}=0\)
\(\Rightarrow a\left(x-1\right)-\dfrac{2\left(x-1\right)}{\sqrt{x}+1}=0\Leftrightarrow\left(x-1\right)\left(a-\dfrac{2}{\sqrt{x}+1}\right)=0\)
\(\Rightarrow a-\dfrac{2}{\sqrt{x}+1}=0\) cũng có nghiệm \(x=1\)
\(\Rightarrow a-\dfrac{2}{1+1}=0\Rightarrow a=1\Rightarrow b=1\)
Thử lại: \(\lim\limits_{x\rightarrow1}\dfrac{x+1-2\sqrt{x}}{\left(x-1\right)^2\left(x+2\right)}=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+2\right)\left(x+1+2\sqrt{x}\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{1}{\left(x+2\right)\left(x+1+2\sqrt{x}\right)}=\dfrac{1}{12}\) (thỏa mãn)
