Đáp án:
$I=2$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right) - 16}}{{x - 1}} = 24\\
\Rightarrow \mathop {\lim }\limits_{x \to 1} \left( {f\left( x \right) - 16} \right) = 0\\
\Rightarrow f\left( 1 \right) - 16 = 0\\
\Rightarrow f\left( 1 \right) = 16
\end{array}$
Lại có:
$\begin{array}{l}
I = \mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right) - 16}}{{\left( {x - 1} \right)\left( {\sqrt {2f\left( x \right) + 4} + 6} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right) - 16}}{{x - 1}}.\mathop {\lim }\limits_{x \to 1} \dfrac{1}{{\sqrt {2f\left( x \right) + 4} + 6}}\\
= 24.\dfrac{1}{{\sqrt {2.16 + 4} + 6}}\\
= 2
\end{array}$
Vậy $I=2$
