Đáp án:
$\begin{array}{l}
\mathop {\lim }\limits_{x \to - 1} f\left( x \right) = \mathop {\lim }\limits_{x \to - 1} \dfrac{{{x^2} + 3x + 2}}{{x + 1}}\\
= \mathop {\lim }\limits_{x \to - 1} \dfrac{{\left( {x + 1} \right)\left( {x + 2} \right)}}{{x + 1}}\\
= \mathop {\lim }\limits_{x \to - 1} \left( {x + 2} \right)\\
= - 1 + 2 = 1\\
\Rightarrow \mathop {\lim }\limits_{x \to - 1} f\left( x \right) \ne f\left( { - 1} \right)
\end{array}$
=> Hàm số đã cho ko liên tục trên R.
