Cho đa thức \(f\left( x \right)\) thỏa mãn \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 4}}{{x - 1}} = 10\). Tính \(L = \mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \rig

Đặt \(g\left( x \right) = \frac{{f\left( x \right) - 4}}{{x - 1}} \Leftrightarrow f\left( x \right) = \left( {x - 1} \right)g\left( x \right) + 4\).

\( \Rightarrow \mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \left( {x - 1} \right)g\left( x \right) + 4 = 4 \Rightarrow f\left( 1 \right) = 4.{\rm{ }}\)

Khi đó: \(L = \mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 4}}{{x - 1}} \cdot \frac{1}{{\left( {x + 1} \right)\left[ {\sqrt {f\left( x \right) + 5} + 3} \right]}} = 10 \cdot \frac{1}{{12}} = \frac{5}{6}\).

Đáp án: \(\frac{5}{6}\).