Cho \(f\left( x \right)\) là đa thức thỏa mãn \(\mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right) - 5}}{{x - 1}} = 10\). Tính \(\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {3f\left( x \right) + 1}  - 4}}{{{x^2} - 1}}\)

Chỉ điền số nguyên, phân số dạng a/b

Đáp án:

Đặt \(g\left( x \right) = \dfrac{{f\left( x \right) - 5}}{{x - 1}}\)

\(\begin{array}{l} \Rightarrow f\left( x \right) = 5 + \left( {x - 1} \right)g\left( x \right)\\ \Rightarrow \mathop {\lim }\limits_{x \to 1} f\left( x \right) = 5\end{array}\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {3f\left( x \right) + 1}  - 4}}{{{x^2} - 1}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{3f\left( x \right) + 1 - 16}}{{\left( {{x^2} - 1} \right)\left[ {\sqrt {3f\left( x \right) + 1}  + 4} \right]}}\\ = \mathop {\lim }\limits_{x \to 1} \left[ {\dfrac{{f\left( x \right) - 5}}{{x - 1}}.\dfrac{3}{{\left( {x + 1} \right)\left[ {\sqrt {3f\left( x \right) + 1}  + 4} \right]}}} \right]\\ = 10.\dfrac{3}{{2.8}} = \dfrac{{15}}{8}\end{array}\)