a) Ta có
$\underset{x \to 2}{\lim} (x+3) = 2 + 3 = 5$
b) Ta có
$\underset{x \to 0}{\lim} (x-3)(x+2) = (0-3)(0+2) = -6$
c) Ta có
$\underset{x \to 2}{\lim} \dfrac{\sqrt{x^2 + 3x -1}}{x-1} = \dfrac{\sqrt{2^2 + 3.2-1}}{2-1} = 3$.
d) Ta có
$\underset{x \to -1}{\lim} \dfrac{x^2-3}{x^3+2} = \dfrac{(-1)^2 - 3}{(-1)^3 + 2} = \dfrac{1-3}{-1+2} = -2$
