Question

Вычислить предел функции

Answer 1

$\displaystyle \lim_{x \to 3}\dfrac{x^2-5x+6}{3x^2-9x}=\lim_{x \to 3}\dfrac{x^2-2x-3x+6}{3x(x-3)}=\lim_{x \to 3}\dfrac{x(x-2)-3(x-2)}{3x(x-3)}=\\ \\ \\ =\lim_{x \to 3}\dfrac{(x-2)(x-3)}{3x(x-3)}=\lim_{x \to 3}\dfrac{x-2}{3x}=\dfrac{3-2}{3\cdot3}=\dfrac{1}{9}$