Giải thích các bước giải:
$\lim_{x\to 3}(\dfrac{1}{x}-\dfrac{1}{3}).\dfrac{1}{(x-3)^3}$
$=\lim_{x\to 3}\dfrac{-(x-3)}{3x}.\dfrac{1}{(x-3)^3}$
$=\lim_{x\to 3}\dfrac{-1}{3x}.\dfrac{1}{(x-3)^2}$
$=\dfrac{-1}{3.3}.\dfrac{1}{(3-3)^2}$
$=\dfrac{-1}{9}.+\infty$
$=-\infty$