Cho các hàm số \(f(x) = \left\{ \begin{array}{l}\frac{{\sqrt {4x - 7}  - 1}}{{{x^2} - 4}} & {\rm{khi}}\,x > 2\\\frac{{5x - 9}}{2} &  & {\rm{khi}}\,x \le 2\end{array} \right.\) và \(g(x) = \left\{ {\begin{array}{*{20}{l}}{\frac{{\sqrt {x + 2}  - 2}}{{2 - x}}}&{{\rm{ khi }}x > 2}\\{\frac{{1 - x}}{4}}&{{\rm{ khi }}x \le 2}\end{array}} \right.\).

Lời giải

a) Đúng  

b) Sai 

c) Sai 

d) Đúng

Ta có: \(f\left( {{x_0}} \right) = f(2) = \frac{1}{2} = \mathop {\lim }\limits_{x \to {2^ - }} f(x)\).

\(\mathop {\lim }\limits_{x \to x_0^ + } f(x) = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{\sqrt {4x - 7}  - 1}}{{{x^2} - 4}} = \frac{1}{2} = \mathop {\lim }\limits_{x \to {2^ - }} f(x)\).

\( \Rightarrow \mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ + }} f(x) = \frac{1}{2} = \mathop {\lim }\limits_{x \to 2} f(x) = f(2)\).

Vậy hàm số \(f\left( x \right)\)liên tục tại điểm \({x_0} = 2\).

Ta có: \(g(2) = \frac{{1 - 2}}{4} =  - \frac{1}{4};\mathop {\lim }\limits_{x \to {2^ - }} g(x) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {\frac{{1 - x}}{4}} \right) =  - \frac{1}{4}\);

\(\mathop {\lim }\limits_{x \to {2^ + }} g(x) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {\frac{{\sqrt {x + 2}  - 2}}{{2 - x}}} \right) = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{x + 2 - 4}}{{(2 - x)(\sqrt {x + 2}  + 2)}} = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{ - 1}}{{\sqrt {x + 2}  + 2}} =  - \frac{1}{4}{\rm{. }}\)

Suy ra \(\mathop {\lim }\limits_{x \to 2} g(x) =  - \frac{1}{4} = g(2)\).

Vậy hàm số \(g(x)\) liên tục tại điểm \({x_0} = 2\).