Tìm giá trị nhỏ nhất của \(a\) để hàm số \(f\left( x \right) = \left\{ {\begin{array}{*{20}{l}}{\dfrac{{{x^2} - 5x + 6}}{{\sqrt {4x - 3}  - x}}}&{{\rm{khi }}x > 3}\\{1 - {a^2}x}&{{\rm{khi }}x \le 3}\end{array}} \right.\) liên tục tại \(x = 3\).

Bước 1:

Ta có:

\(f\left( 3 \right) = 1 - 3{a^2}\)

\(\mathop {\lim }\limits_{x \to {3^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{{x^2} - 5x + 6}}{{\sqrt {4x - 3}  - x}}\) \( = \mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{\left( {x - 2} \right)\left( {x - 3} \right)\left( {\sqrt {4x - 3}  + x} \right)}}{{\left( {\sqrt {4x - 3}  - x} \right)\left( {\sqrt {4x - 3}  + x} \right)}}\) \( = \mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{\left( {x - 2} \right)\left( {x - 3} \right)\left( {\sqrt {4x - 3}  + x} \right)}}{{4x - 3 - {x^2}}}\) \( = \mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{\left( {x - 2} \right)\left( {x - 3} \right)\left( {\sqrt {4x - 3}  + x} \right)}}{{ - \left( {x - 3} \right)\left( {x - 1} \right)}}\)

\( = \mathop {\lim }\limits_{x \to {3^ + }} \dfrac{{\left( {x - 2} \right)\left( {\sqrt {4x - 3}  + x} \right)}}{{ - \left( {x - 1} \right)}}\) \( = \dfrac{{\left( {3 - 2} \right)\left( {\sqrt {12 - 3}  + 3} \right)}}{{ - \left( {3 - 1} \right)}} =  - 3\)

\(\mathop {\lim }\limits_{x \to {3^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {3^ - }} \left( {1 - {a^2}x} \right) = 1 - 3{a^2}\)

Bước 2:

Do đó hàm số liên tục tại \(x = 3\) \( \Leftrightarrow \mathop {\lim }\limits_{x \to {3^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {3^ - }} f\left( x \right) = f\left( 3 \right)\)

\( \Leftrightarrow 1 - 3{a^2} =  - 3 \Leftrightarrow 3{a^2} = 4\) \( \Leftrightarrow {a^2} = \dfrac{4}{3} \Leftrightarrow a =  \pm \dfrac{2}{{\sqrt 3 }}\)

\(\Rightarrow {a_{\min }} =  - \dfrac{2}{{\sqrt 3 }}.\)