Cho đa thức f(x) thỏa mãn 

Bước 1:

Đặt\[g\left( x \right) = \frac{{f\left( x \right) - 2}}{{x - 1}} \Rightarrow f\left( x \right) = \left( {x - 1} \right)g\left( x \right) + 2\]

\[ \Rightarrow \mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \left[ {\left( {x - 1} \right)g\left( x \right) + 2} \right] = 2\]

Bước 2:

Ta có:

\[\begin{array}{*{20}{l}}{\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 2}}{{\left( {{x^2} - 1} \right)\left[ {f\left( x \right) + 1} \right]}}}\\{ = \mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 2}}{{x - 1}}.\frac{1}{{\left( {x + 1} \right)\left[ {f\left( x \right) + 1} \right]}}}\\{ = 12.\frac{1}{{2.\left( {2 + 1} \right)}} = 2}\end{array}\]