Cho hàm số \[f(x)\] xác định với mọi \[x \ne 0\] thỏa mãn \[f(x) + 3f\left( {\frac{1}{x}} \right) = 8x,\,\,x \ne 0\]. Tính \[\mathop {\lim }\limits_{x \to \sqrt 3 } \frac{{f(x)}}{{x - \sqrt 3 }}\]

Chọn B

Từ \[f(x) + 3f\left( {\frac{1}{x}} \right) = 8x,\,\,x \ne 0\], thay \[x\] bởi \[\frac{1}{x}\] ta được \[f\left( {\frac{1}{x}} \right) + 3f\left( x \right) = \frac{8}{x}\]. Suy ra

\[\left\{ \begin{array}{l}f\left( {\frac{1}{x}} \right) + 3f\left( x \right) = \frac{8}{x}\\f\left( x \right) + 3f\left( {\frac{1}{x}} \right) = 8x\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}3f\left( {\frac{1}{x}} \right) + 9f\left( x \right) = \frac{{24}}{x}\\f\left( x \right) + 3f\left( {\frac{1}{x}} \right) = 8x\end{array} \right. \Rightarrow 8f\left( x \right) = 8\left( {\frac{3}{x} - x} \right)\]

\[ \Rightarrow f\left( x \right) = \frac{3}{x} - x = \frac{{3 - {x^2}}}{x}\]. Do đó

\[\mathop {\lim }\limits_{x \to \sqrt 3 } \frac{{f(x)}}{{x - \sqrt 3 }} = \mathop {\lim }\limits_{x \to \sqrt 3 } \frac{{3 - {x^2}}}{{x\left( {x - \sqrt 3 } \right)}} = \mathop {\lim }\limits_{x \to \sqrt 3 } \frac{{ - \left( {\sqrt 3 + x} \right)}}{x} = - 2\].