Cho hàm số \(f\left( x \right) = \left\{ \begin{array}{l}\frac{{\sqrt {x + 3} - 2}}{{{x^2} - 1}},x > 1\\ax + 2,\,\,\,\,\,\,\,\,\,\,\,x \le 1\end{array} \right.\). Giá trị của \(a\...

* Hướng dẫn giải

Ta có: \(f\left( 1 \right) = a + 2\).

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

+) \(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{\sqrt {x + 3}  - 2}}{{{x^2} - 1}}\)\( = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{x + 3 - 4}}{{\left( {{x^2} - 1} \right)\left( {\sqrt {x + 3}  + 2} \right)}}\) \( = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{x - 1}}{{\left( {{x^2} - 1} \right)\left( {\sqrt {x + 3}  + 2} \right)}}\) \( = \mathop {\lim }\limits_{x \to {1^ + }} \frac{1}{{\left( {x + 1} \right)\left( {\sqrt {x + 3}  + 2} \right)}}\) \( = \frac{1}{{\left( {1 + 1} \right)\left( {\sqrt {1 + 3}  + 2} \right)}} = \frac{1}{8}\)

\( \Rightarrow \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \frac{1}{8}\)

Hàm số liên tục tại \(x = 1\) \( \Leftrightarrow \mathop {\lim }\limits_{x \to 1} f\left( x \right) = f\left( 1 \right)\)\( \Leftrightarrow a + 2 = \frac{1}{8}\) \( \Leftrightarrow a =  - \frac{{15}}{8}\).