Câu hỏi
Cho hàm số \(f\left( x \right) = \left\{ \begin{array}{l}\dfrac{{2 - \sqrt {4 - x} }}{x}\,\,khi\,\,x \ne 0\\\,\,\,\,\,\,\,\dfrac{1}{4}\,\,\,\,\,\,\,\,\,\,\,\,\,khi\,\,x = 0\end{array} \right.\). Khi đó \(f'\left( 0 \right)\) là kết quả nào trong các kết quả sau:
- A không tồn tại \(f'\left( 0 \right)\).
- B \(\dfrac{1}{{32}}.\)
- C \(\dfrac{1}{{64}}.\)
- D \(\dfrac{1}{4}.\).
Phương pháp giải:
- Xét tính liên tục của hàm số tại \(x = 0\).
- Tính giới hạn \(\mathop {\lim }\limits_{x \to 0} \dfrac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}}\). Nếu giới hạn này tồn tại thì \(f'\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}}\).
Lời giải chi tiết:
Ta có:
\(\begin{array}{l}\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{2 - \sqrt {4 - x} }}{x} = \mathop {\lim }\limits_{x \to 0} \dfrac{{4 - 4 + x}}{{x\left( {2 + \sqrt {4 - x} } \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} \dfrac{1}{{2 + \sqrt {4 - x} }} = \dfrac{1}{{2 + 2}} = \dfrac{1}{4}\\ \Rightarrow \mathop {\lim }\limits_{x \to 0} f\left( x \right) = f\left( 0 \right)\end{array}\)
\( \Rightarrow \) Hàm số đã cho liên tục tại \(x = 0\).
Ta có:
\(\begin{array}{l}\mathop {\lim }\limits_{x \to 0} \dfrac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{2 - \sqrt {4 - x} }}{x} - \dfrac{1}{4}}}{x}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} \dfrac{{8 - 4\sqrt {4 - x} - x}}{{4{x^2}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\left( {8 - x} \right)}^2} - 16\left( {4 - x} \right)}}{{4{x^2}\left( {8 - x + 4\sqrt {4 - x} } \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2} - 16x + 64 - 64 + 16x}}{{4{x^2}\left( {8 - x + 4\sqrt {4 - x} } \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} \dfrac{1}{{4\left( {8 - x + 4\sqrt {4 - x} } \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{1}{{4\left( {8 - 0 + 4.2} \right)}} = \dfrac{1}{{64}}\end{array}\)
Vậy \(f'\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \dfrac{1}{{64}}\).
Chọn C.
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