Đáp án:
$\begin{array}{l}
\mathop {\lim }\limits_{x \to - {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to - {1^ + }} \left( { - 1} \right) = - 1\\
\mathop {\lim }\limits_{x \to - {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to - {1^ - }} \frac{{{x^2} + x}}{{x + 1}} = \mathop {\lim }\limits_{x \to - {1^ - }} \frac{{x\left( {x + 1} \right)}}{{x + 1}} = \mathop {\lim }\limits_{x \to - {1^ - }} x = - 1\\
\Rightarrow \mathop {\lim }\limits_{x \to - {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to - {1^ - }} f\left( x \right) = - 1
\end{array}$
Vậy có tồn tại $\mathop {\lim }\limits_{x \to - 1} f\left( x \right)$
