Cho hàm số \(f\left( x \right) = \left\{ \begin{array}{l}\frac{1}{4}x + \frac{1}{4}\;\;\;\;\;\;\;\;{\rm{khi}}\;\;x \le 2\\\frac{{\sqrt {3x - 2} - 2}}{{x - 2}}\;\;{\rm{khi}}\;\;x > 2\end{array} \right.\).

a) \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{\sqrt {3x - 2} - 2}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{3\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {\sqrt {3x - 2} + 2} \right)}}\)\( = \mathop {\lim }\limits_{x \to {2^ + }} \frac{3}{{\sqrt {3x - 2} + 2}} = \frac{3}{4}\).

b) \(\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{4}x + \frac{1}{4}} \right)\)\( = \frac{1}{4}\).

c) \(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {\frac{1}{4}x + \frac{1}{4}} \right) = \frac{3}{4}\).

Có \(f\left( 2 \right) = \frac{3}{4}\).

Vì \(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = f\left( 2 \right)\). Do đó hàm số liên tục tại \(x = 2\).

d) \(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {\frac{1}{4}x + \frac{1}{4}} \right) = \frac{3}{4}\)

Đáp án: a) Sai; b) Đúng; c) Đúng; d) Đúng.