b) Ta có
$\underset{x \to 1}{\lim} f(x) = \underset{x \to 1}{\lim} \dfrac{\sqrt{x+3} - 2}{x-1}$
$= \underset{x \to 1}{\lim} \dfrac{x + 3 - 4}{(x-1)(\sqrt{x+3} + 2)}$
$= \underset{x \to 1}{\lim} \dfrac{x-1}{(x-1)(\sqrt{x+3} + 2)}$
$= \underset{x \to 1}{\lim} \dfrac{1}{\sqrt{x+3} + 2}$
$= \dfrac{1}{\sqrt{1 + 3} + 2} = \dfrac{1}{4} = f(1)$
Ta thấy rằng
$\underset{x \to 1}{\lim} f(x) = f(1)$
Vậy hso liên tục tại $x = 1$.
c) Ta có
$\underset{x \to 2}{\lim} f(x) = \underset{x \to 2}{\lim} \dfrac{4-x^2}{x^2 - 3x + 2}$
$= \underset{x \to 2}{\lim} \dfrac{(2-x)(2+x)}{(x-1)(x-2)}$
$= \underset{x \to 2}{\lim} - \dfrac{x + 2}{x-1}$
$= - \dfrac{2 + 2}{2 - 1} = -4 \neq f(2)$
Ta thấy rằng
$\underset{x \to 2}{\lim} f(x) \neq f(2)$
Vậy hso ko liên tục tại $x = 2$.
