Cho hàm số f(x) = {3 - căn {x + 9} / x

Xét tại điểm \({x_0} = 0\)

\(\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{3 - \sqrt {x + 9} }}{x} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{ - x}}{{x(3 + \sqrt {x + 9} )}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{ - 1}}{{(3 + \sqrt {x + 9} )}} = \frac{{ - 1}}{6}\)

\(\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} \left( {x - \frac{1}{6}} \right) = \frac{{ - 1}}{6}\)

\(f(0) = \frac{{ - 1}}{6}\)

\( \Rightarrow f(0) = \mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} f(x)\) nên hàm số liên tục tại\({x_0} = 0\)