Bài tập 33 trang 159 SGK Toán 11 NC

Ta có:

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {4x - 3} \right)\\
 = 4.2 - 3 = 5
\end{array}\\
\begin{array}{l}
\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {{x^2} - 2x + 3} \right)\\
 = {2^2} - 2.2 + 3 = 3
\end{array}
\end{array}\)

Vì \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right)\) nên không tồn tại \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\).

-- Mod Toán 11 HỌC247