Đáp án:
$\begin{array}{l}
2)\\
+ )\mathop {\lim }\limits_{x \to - 2} f\left( x \right) = \mathop {\lim }\limits_{x \to - 2} \frac{{x + 2}}{{2{x^2} - 8}}\\
= \mathop {\lim }\limits_{x \to - 2} \frac{{x + 2}}{{2\left( {x + 2} \right)\left( {x - 2} \right)}}\\
= \mathop {\lim }\limits_{x \to - 2} \frac{1}{{2\left( {x - 2} \right)}} = \frac{{ - 1}}{8}\\
+ )f\left( { - 2} \right) = \frac{{ - 1}}{6} \ne \mathop {\lim }\limits_{x \to - 2} f\left( x \right)
\end{array}$
Vậy hs ko liên tục tại x=-2
3)
$\begin{array}{l}
+ \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {2{x^2} - x} \right) = 2.1 - 1 = 1\\
+ \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( {2 - x} \right) = 2 - 1 = 1\\
\Rightarrow \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = 1
\end{array}$
Vậy hs liên tục trên R.
