Cho hàm số \(f\left( x \right) = \left\{ \begin{array}{l}
\frac{{\sqrt {3x + 1} - 2x}}{{x - 1}}\,\,\,khi\,\,x \ne 1\\
- \frac{5}{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,khi\,\,x = 1
\end{array} \right.\). Tính \(f'(1)\)
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Lời giải:
Báo saiTrước hết ta xét tính liên tục của hàm số tại x = 1
Ta có
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {3x + 1} - 2x}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt {3x + 1} - 2x} \right)\left( {\sqrt {3x + 1} + 2x} \right)}}{{\left( {x - 1} \right)\left( {\sqrt {3x + 1} + 2x} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{3x + 1 - 4{x^2}}}{{\left( {x - 1} \right)\left( {\sqrt {3x + 1} + 2x} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{ - \left( {x - 1} \right)\left( {4x + 1} \right)}}{{\left( {x - 1} \right)\left( {\sqrt {3x + 1} + 2x} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{ - 4x - 1}}{{\sqrt {3x + 1} + 2x}} = \frac{{ - 4 - 1}}{{\sqrt 4 + 2}} = \frac{{ - 5}}{4} = f\left( 1 \right)
\end{array}\)
\( \Rightarrow \) Hàm số liên tục tại x = 1
Tính \(f'(1)\)
\(\begin{array}{l}
\Rightarrow f'\left( 1 \right) = \mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}} = \frac{{\frac{{\sqrt {3x + 1} - 2x}}{{x - 1}} + \frac{5}{4}}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{4\sqrt {3x + 1} - 8x + 5x - 5}}{{4{{\left( {x - 1} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{4\sqrt {3x + 1} - 3x - 5}}{{4{{\left( {x - 1} \right)}^2}}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {4\sqrt {3x + 1} - 3x - 5} \right)\left( {4\sqrt {3x + 1} + 3x + 5} \right)}}{{4{{\left( {x - 1} \right)}^2}\left( {4\sqrt {3x + 1} + 3x + 5} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{16\left( {3x + 1} \right) - \left( {9{x^2} + 30x + 25} \right)}}{{4{{\left( {x - 1} \right)}^2}\left( {4\sqrt {3x + 1} + 3x + 5} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{ - 9{x^2} + 18x - 9}}{{4{{\left( {x - 1} \right)}^2}\left( {4\sqrt {3x + 1} + 3x + 5} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{ - 9{{\left( {x - 1} \right)}^2}}}{{4{{\left( {x - 1} \right)}^2}\left( {4\sqrt {3x + 1} + 3x + 5} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{ - 9}}{{4\left( {4\sqrt {3x + 1} + 3x + 5} \right)}} = \frac{{ - 9}}{{64}}
\end{array}\)
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