For example, a case of "0/0":
sin(x) cos(x) 1 lim ------ = lim ------ = --- = 1 x→0 x x→0 1 1and a case of "∞/∞":
√x 1/(2√x) √x lim ----- = lim ------- = lim ---- = ∞ x→∞ ln(x) x→∞ 1/x x→∞ 2Sometimes, even limits which don't appear to be quotients can be handled with the same rule:
(x + √(x2 - x) (x - √(x2 - x) x2 - (x2 - x) lim x - √(x2 - x) = lim ----------------------------- = lim -------------- x→∞ x→∞ (x + √(x2 + x) x→∞ x + √(x2 + x
x 1 1 = lim -------------- = lim ------------------------ = ---- = 1/2 x→∞ (x + √(x2 + x) x→∞ 1 + (2x + 1)/(2√(x2 + x)) 1+1