lim $\frac{x-√(3x - 2)}{x² - 4}$ = lim $\frac{x²-3x +2}{(x+2)(x-2)[x+√(3x - 2)]}$
x->2 x->2
= lim $\frac{(x-2)(x-1)}{(x+2)(x-2)[x+√(3x - 2)]}$
x->2
= lim $\frac{x-1}{(x+2)[x+√(3x - 2)]}$
x->2
= $\frac{2-1}{(2+2)[2+√(3.2 - 2)]}$
= $\frac{1}{16}$
Vậy lim $\frac{x-√(3x - 2)}{x² - 4}$ = $\frac{1}{16}$
x->2