Tìm \(m\) để hàm số \[f(x) = \left\{ \begin{array}{l}\frac{{{x^2} + 4x + 3}}{{x + 1}}\,\,\,khi\,\,x > - 1\\mx + 2\,\,\,\,\,\,\,\,\,\,\,\,\,khi\,\,x \le - 1\end{array} \right.\] liên tục tại điểm \(x = - 1\).

Ta có: \(\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} f\left( x \right)\)\( = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \frac{{{x^2} + 4x + 3}}{{x + 1}}\)\( = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \frac{{\left( {x + 1} \right)\left( {x + 3} \right)}}{{x + 1}}\)\( = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \left( {x + 3} \right)\)\( = 2\).

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

\(f\left( { - 1} \right) =  - m + 2\).

Để hàm số đã cho liên tục tại điểm \(x =  - 1\) thì \(\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} f\left( x \right) = f\left( { - 1} \right)\)\( \Leftrightarrow 2 =  - m + 2\)\( \Leftrightarrow m = 0\).