#y=f(x)=k/g(x)#
Existence Condition is:
#g(x)!=0#
#:.x-9!=0#
#:.x!=9#
Then:
#F.E.#= Field of Existence=Domain: #x in (-oo,9)uu(9,+oo)#
#x=9# could be a vertical asymptote
To find the range we have to study the behavior for:
#lim_(x rarr -oo) f(x)=lim_(x rarr -oo) 5/(x-9)=5/-oo=0^-#
#lim_(x rarr +oo) f(x)=lim_(x rarr +oo) 5/(x-9)=5/(+oo)=0^+#
Then
#y=0# is a horizontal asymptote.
Indeed,
#f(x)!=0 AAx in F.E.#
#lim_(x rarr 9^-) f(x)=lim_(x rarr 9^-) 5/(x-9)=5/0^(-)=-oo#
#lim_(x rarr 9^+) f(x)=lim_(x rarr 9^+) 5/(x-9)=5/0^(+)=+oo#
Then
#x=9# it's a vertical asympote
#:. # Range of #f(x)#: #y in (-oo,0)uu(0,+oo)#