Question

Вычислить предел функции

Answer 1

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