How do you evaluate the limit of #lim (x^3-x^2+2x)/(x^3+4x^2-2x)# as #x->0#?
1 Answer
Dec 13, 2016
Explanation:
#(x^3-x^2+2x)/(x^3+4x^2-2x) = ((x^2-x+2)color(red)(cancel(color(black)(x))))/((x^2+4x-2)color(red)(cancel(color(black)(x)))) = (x^2-x+2)/(x^2+4x-2)#
So:
#lim_(x->0) (x^3-x^2+2x)/(x^3+4x^2-2x) = lim_(x->0) (x^2-x+2)/(x^2+4x-2) = (0+0+2)/(0+0-2) = -1#