Cho đa thức \(f\left( x \right)\) thỏa mãn \(\mathop {\lim }\limits_{x \to 4} \dfrac{{f\left( x \right) - 2018}}{{x - 4}} = 2019\). Biết \(L = \mathop {\lim }\limits_{x \to 4} \dfrac{{1009\left[ {f\left( x \right) - 2018} \right]}}{{\left( {\sqrt x  - 2} \right)\left[ {\sqrt {2019f\left( x \right) + 2019}  + 2019} \right]}}\).

Đáp án: $L=$

Bước 1: Tính $\mathop {\lim }\limits_{x \to 4} f\left( x \right)$

Đặt \(\dfrac{{f\left( x \right) - 2018}}{{x - 4}} = g\left( x \right) \Rightarrow f\left( x \right) = \left( {x - 4} \right)g\left( x \right) + 2018\)

\( \Rightarrow \mathop {\lim }\limits_{x \to 4} f\left( x \right) = 2018\).

Bước 2: Nhân cả tử và mẫu với $\sqrt{x}+2$. Tính $L$

\(\begin{array}{l}L = \mathop {\lim }\limits_{x \to 4} \dfrac{{1009\left[ {f\left( x \right) - 2018} \right]}}{{\left( {\sqrt x  - 2} \right)\left[ {\sqrt {2019f\left( x \right) + 2019}  + 2019} \right]}}\\\,\,\,\, = \mathop {\lim }\limits_{x \to 4} \dfrac{{1009\left[ {f\left( x \right) - 2018} \right]\left( {\sqrt x  + 2} \right)}}{{\left( {x - 4} \right)\left[ {\sqrt {2019f\left( x \right) + 2019}  + 2019} \right]}}\\\,\,\,\, = 1009.\mathop {\lim }\limits_{x \to 4} \dfrac{{f\left( x \right) - 2018}}{{x - 4}}.\dfrac{{\sqrt x  + 2}}{{\sqrt {2019f\left( x \right) + 2019}  + 2019}}\\\,\,\,\, = 1009.2019.\dfrac{{\sqrt {2018}  + 2}}{{\sqrt {2019.2018 + 2019}  + 2019}}\\\,\,\,\, = 1009.2019.\dfrac{{\sqrt 4  + 2}}{{2019 + 2019}} = 2018\end{array}\)