Question 937818
limit as x approaches 0 of ((x-2)^3 + 8)/x is 12
(x-2)^3 = (x^3 -6x^2 +12x - 8) = x((x-6)x+12) -8
above result comes from factoring x
now we can return to our problem
limit as x approaches 0 of ((x-2)^3 + 8)/x = (x((x-6)x+12) -8 +8) / x =  (x((x-6)x+12)) / x = ((x-6)x+12) = 12 as x approaches 0
above result comes from canceling x from numerator and denominator