Cho hàm số \(f\left( x \right) = \left\{ \begin{array}{l}\frac{{3 - \sqrt {4 - x} }}{4}{\mkern 1mu} {\mkern 1mu} khi{\mkern 1mu} {\mkern 1mu} x \n

Câu hỏi :

Cho hàm số \(f\left( x \right) = \left\{ \begin{array}{l}
\frac{{3 - \sqrt {4 - x} }}{4}{\mkern 1mu} {\mkern 1mu} khi{\mkern 1mu} {\mkern 1mu} x \ne 0\\
\frac{1}{4}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} khi{\mkern 1mu} {\mkern 1mu} x = 0
\end{array} \right..\) Khi đó \(f'\left( 0 \right)\) là kết quả nào sau đây?

A. \(\frac{1}{4}\)

B. \(\frac{1}{{16}}\)

C. \(\frac{1}{{32}}\)

D. Không tồn tại 

* Đáp án

B

* Hướng dẫn giải

Ta có \(\mathop {\lim }\limits_{x \to {x_0}} {\mkern 1mu} \frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to 0} {\mkern 1mu} \frac{{\frac{{3 - \sqrt {4 - x} }}{4} - \frac{1}{4}}}{{x - 0}} = \mathop {\lim }\limits_{x \to 0} {\mkern 1mu} \frac{{2 - \sqrt {4 - x} }}{{4x}} = \mathop {\lim }\limits_{x \to 0} {\mkern 1mu} \frac{1}{{4\left( {2 + \sqrt {4 - x} } \right)}} = \frac{1}{{16}}\)

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