Substituting #3# in the given expression, to find its limit, results to
#" "#
an indeterminate solution :
#lim_(x->3)abs(x^2-9)/abs(x-3)#
#=(abs(3^2-9))/(abs(3-3))#
#=0/0# #" "# Indeterminate solution.
#""#
Factorizing #" "x^2-9" "# is the ideal way to find the solution.
#x^2-9=x^2-3^2=(x-3)(x+3)#
#" "#
#" "#
#lim_(x->3^-)abs(x^2-9)/abs(x-3)=lim_(x->3^-)abs((x-3)(x+3))/abs(x-3)#
#" "#
#=lim_(x->3^-)(abs(x-3)abs(x+3))/abs(x-3)" "
=lim_(x->3^-)((3-x)(x+3))/(3-x)#
#" "#
#=lim_(x->3^-)(cancel((3-x))(x+3))/cancel((3-x))=lim_(x->3^-)3+x=3+3=6#
#" "#
#" "#
#" "#
#lim_(x->3^+)abs(x^2-9)/abs(x-3)=lim_(x->3^+)abs((x-3)(x+3))/abs(x-3)#
#" "#
#=lim_(x->3^+)(abs(x-3)abs(x+3))/abs(x-3)" "
=lim_(x->3^+)((x-3)(x+3))/(x-3)#
#" "#
#=lim_(x->3^+)(cancel((x-3))(x+3))/cancel((x-3))=lim_(x->3^+)x+3=3+3=6#
#" "#
#" "#
Therefore,
#" "#
#lim_(x->3)abs(x^2-9)/abs(x-3)=lim_(x->3^-)abs(x^2-9)/abs(x-3)=lim_(x->3^+)abs(x^2-9)/abs(x-3)=6#
