How do you evaluate the limit of #lim ((x+2)^2-4)/x# as #x->0#?
1 Answer
Nov 18, 2016
Explanation:
The function evaluated at x = 0 is
#0/0# that is indeterminate.Expand the numerator.
#(x^2+4x+4-4)/x=(x^2+4x)/x=x+4#
#lim_(xto0)((x+2)^2-4)/x=lim_(xto0)(x+4)=4#
