Question

[Dang Tung]

 

Answer 1

[Nguyễn Việt Lâm]

Do \(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-1}{x^2-3x+2}=1\) hữu hạn nên \(f\left(x\right)-1=0\) có nghiệm \(x=1\)

\(\Rightarrow f\left(1\right)=1\)

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

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

\(=1.\dfrac{2-1}{1+1}+1.\dfrac{2-1}{\left(1+1\right).\left(1+\sqrt{2-f\left(1\right)}\right)}=\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{3}{4}\)

Cách 2: \(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-1}{x^2-3x+2}=1\Rightarrow\) chọn \(f\left(x\right)=2-x\)

Khi đó:

\(\lim\limits_{x\rightarrow1}\dfrac{2-x-\sqrt{x}}{1-x^2}=\dfrac{3}{4}\)