Cho hàm số \(f\left( x \right) = \left\{ \begin{array}{l}3x - 2\,\,\,khi\,\,x \ge 1\\m{x^2} - mx + 1\,\,\,khi\,\,x...

* Hướng dẫn giải

Hàm số \(y=f(x)\) liên tục tại điểm \(x = 1 \Leftrightarrow \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = f\left( 1 \right)\) 

Ta có: \(f\left( 1 \right) = 3.1 - 2 = 1\) 

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