Đáp án:
\[f'\left( 0 \right) = \frac{1}{4}\]
Giải thích các bước giải:
\(\begin{array}{l}
f'\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \frac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to 0} \frac{{3 - \sqrt {4 - x} - 1}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{2 - \sqrt {4 - x} }}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {2 - \sqrt {4 - x} } \right)\left( {2 + \sqrt {4 - x} } \right)}}{{x\left( {2 + \sqrt {4 - x} } \right)}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{{2^2} - \left( {4 - x} \right)}}{{x\left( {2 + \sqrt {4 - x} } \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{x}{{x\left( {2 + \sqrt {4 - x} } \right)}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{1}{{2 + \sqrt {4 - x} }} = \frac{1}{{2 + \sqrt {4 - 0} }} = \frac{1}{4}
\end{array}\)
Vậy \(f'\left( 0 \right) = \frac{1}{4}\)
