Cho đa thức \(f\left( x \right)\) là thỏa mãn \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 16}}{{x - 1}} = 24\). Tính \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \rig
Giải thích
Vì \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 16}}{{x - 1}} = 24 \Rightarrow f\left( 1 \right) = 16\) vì nếu \(f(1) \ne 16\) thì \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 16}}{{x - 1}} = \infty \).
Ta có \(I = \mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 16}}{{\left( {x - 1} \right)\left[ {\sqrt {2f\left( x \right) + 4} + 6} \right]}} = \frac{1}{{12}}\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 16}}{{x - 1}} = 2\). Đáp án: 2.