Tìm tham số a để hàm số \(f\left( x \right) = \left\{ \begin{array}{l}\frac{{\sqrt {{x^2} + 5} - 3}}{{x + 2}}\,\,\,khi\,\,\,x \ne - 2\\ax + 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,khi\,\,\,x = - 2\end{array} \right.\) liên tục tại \({x_0} = - 2\)

Ta có:

\(\begin{array}{l}\mathop {\lim }\limits_{x \to  - 2} f\left( x \right) = \mathop {\lim }\limits_{x \to  - 2} \frac{{\sqrt {{x^2} + 5}  - 3}}{{x + 2}}\\ = \mathop {\lim }\limits_{x \to  - 2} \frac{{{x^2} + 5 - 9}}{{\left( {x + 2} \right)\left( {\sqrt {{x^2} + 5}  + 3} \right)}}\\ = \mathop {\lim }\limits_{x \to  - 2} \frac{{{x^2} - 4}}{{\left( {x + 2} \right)\left( {\sqrt {{x^2} + 5}  + 3} \right)}}\\ = \mathop {\lim }\limits_{x \to  - 2} \frac{{\left( {x - 2} \right)\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {\sqrt {{x^2} + 5}  + 3} \right)}}\\ = \mathop {\lim }\limits_{x \to  - 2} \frac{{x - 2}}{{\sqrt {{x^2} + 5}  + 3}}\\ = \frac{{ - 2 - 2}}{{\sqrt {{{\left( { - 2} \right)}^2} + 5}  + 3}}\\ =  - \frac{2}{3}\\f\left( { - 2} \right) =  - 2a + 1\end{array}\)

Hàm số liên tục tại \(x =  - 2\)

\(\begin{array}{l} \Leftrightarrow \mathop {\lim }\limits_{x \to  - 2} f\left( x \right) = f\left( { - 2} \right)\\ \Leftrightarrow  - \frac{2}{3} =  - 2a + 1\\ \Leftrightarrow  - 2a =  - \frac{5}{3}\\ \Leftrightarrow a = \frac{5}{6}\end{array}\)