Cho các hàm số f ( x ) = ⎧ ⎨ ⎩ √ 4 x − 7 − 1 x 2 − 4 k h i x > 2 5 x − 9 2 k h i x ≤ 2 và g ( x ) = ⎧ ⎨ ⎩ √ x + 2 − 2 2 − x k h i x > 2 1 − x 4 k h i x ≤ 2 .

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 \({x_0} = 2\).