Phương pháp giải:

\(f\left( x \right)\) liên tục tại \({x_0}\)\( \Leftrightarrow f\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to x_0^ + } f\left( x \right) = \mathop {\lim }\limits_{x \to x_0^ - } f\left( x \right)\)

Lời giải chi tiết:

\(\begin{array}{l}f\left( 2 \right) = \mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = m + 1\\\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {{x^2} - 1} \right) = 3\\ \Rightarrow m + 1 = 3 \Leftrightarrow m = 2\end{array}\)