Đáp án:
\[f'\left( 0 \right) = \frac{1}{{64}}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
f'\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \frac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{\frac{{2 - \sqrt {4 - x} }}{x} - \frac{1}{4}}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{\frac{{8 - 4\sqrt {4 - x} - x}}{{4x}}}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{8 - 4\sqrt {4 - x} - x}}{{4{x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\left( {4 - x} \right) - 4\sqrt {4 - x} + 4}}{{4{x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {\sqrt {4 - x} - 2} \right)}^2}}}{{4{x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} {\left[ {\frac{{\sqrt {4 - x} - 2}}{{2x}}} \right]^2}\\
= \mathop {\lim }\limits_{x \to 0} {\left[ {\frac{{\left( {\sqrt {4 - x} - 2} \right)\left( {\sqrt {4 - x} + 2} \right)}}{{2x\left( {\sqrt {4 - x} + 2} \right)}}} \right]^2}\\
= \mathop {\lim }\limits_{x \to 0} {\left[ {\frac{{\left( {4 - x} \right) - {2^2}}}{{2x\left( {\sqrt {4 - x} + 2} \right)}}} \right]^2}\\
= \mathop {\lim }\limits_{x \to 0} {\left[ {\frac{{ - x}}{{2x\left( {\sqrt {4 - x} + 2} \right)}}} \right]^2}\\
= \mathop {\lim }\limits_{x \to 0} {\left[ {\frac{{ - 1}}{{2\left( {\sqrt {4 - x} + 2} \right)}}} \right]^2}\\
= {\left[ {\frac{{ - 1}}{{2.\left( {\sqrt {4 - 0} + 2} \right)}}} \right]^2} = \frac{1}{{64}}
\end{array}\)
Vậy \(f'\left( 0 \right) = \frac{1}{{64}}\)
