Ta có
$\underset{x \to 2}{\lim} f(x) = \underset{x \to 2}{\lim} \dfrac{\sqrt{x+2} - 2}{x-2}$
$= \underset{x \to 2}{\lim} \dfrac{x + 2 - 4}{(x-2)(\sqrt{x + 2} + 2)}$
$= \underset{x \to 2}{\lim} \dfrac{1}{\sqrt{x+2} + 2}$
$= \dfrac{1}{2 + 2} = \dfrac{1}{4}$
Lại có
$f(2) = 2.2-3 = 1$
Ta thấy
$\underset{x \to 2}{\lim} f(x) \neq f(2)$
Vậy hso ko liên tục tại $x = 2$.
Bài 3
Ta có
$\underset{x \to -3^+}{\lim} f(x) = \underset{x \to -3^+}{\lim} \dfrac{4x^2 + 9x - 9}{\sqrt{4 - 7x} - 5}$
$= \underset{x \to -3^+}{\lim} \dfrac{(x+3)(4x -3)(\sqrt{4 - 7x} + 5)}{4 - 7x - 25}$
$= \underset{x \to -3^+}{\lim} \dfrac{(x+3)(4x-3)(\sqrt{4-7x} + 5)}{-7(x + 3)}$
$= \underset{x \to -3^+}{\lim} \dfrac{(4x-3)(\sqrt{4-7x} + 5)}{-7}$
$= \dfrac{[4(-3) - 3](\sqrt{4 + 21} + 5)}{-7}$
$= \dfrac{-15 . 10}{-7} = \dfrac{150}{7}$
Lại có
$\underset{x \to -3^-}{\lim} f(x) = \underset{x \to -3^-}{\lim} \dfrac{5x+a}{x-4}$
$= \dfrac{-15 + a}{-3-4}$
$= \dfrac{a-15}{-7} = \dfrac{15-a}{7}$
Để hso có gới hạn tại $x = -3$ thì
$\underset{x \to -3^-}{\lim} f(x) = \underset{x \to -3^+}{\lim} f(x)$
$<-> \dfrac{15-a}{7} = \dfrac{150}{7}$
$<-> 15-a = 150$
$<-> a = -135$
Vậy $a = -135$.
