\(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-1}{\left(x-1\right)\left(x+2\right)}\) hữu hạn \(\Rightarrow f\left(x\right)-1=0\) có nghiệm \(x=1\Rightarrow f\left(1\right)=1\)
Mặt khác \(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-1}{\left(x-1\right)\left(x+2\right)}=\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-1}{\left(x-1\right).\left(1+2\right)}=\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-1}{3\left(x-1\right)}\)
\(\Rightarrow\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-1}{x-1}=9\)
Do đó:
\(\lim\limits_{x\rightarrow1}\dfrac{f^3\left(x\right)+2f\left(x\right)-3}{x^2-x}=\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-1}{x-1}.\dfrac{f^2\left(x\right)+f\left(x\right)+3}{x}=9.\dfrac{1^2+1+3}{1}=45\)
