#f(x)=log(g(x))#
The Existence Condition is
#g(x)>0#
because #log# is definited #AAx in (0,+oo)#
#g(x)=x+2#
#x+2>0#
#x> -2#
Then:
#F.E.# (Field of Existence): #(-2,+oo)#
#x=x_0=-2#
Could be a vertical asymptote if
#lim_(x rarr-2^+) f(x)=+-oo#
#lim_(x rarr-2^+) f(x)=lim_(x rarr-2^+) log(x+2)=#
#lim_(x rarr-2^+) log(0^+)=-oo#
#:. x=-2# vertical asymptote
We could looking for horizontal/slant asymptotes
#lim_(x rarr +oo) f(x)=lim_(x rarr +oo)log(x+2)=+oo#
#:.# no horizontal asymptotes
the slant asymptote formula is
#y=mx+q#
with
#m=lim_(x rarr +oo)f(x)/x#
#q=lim_(x rarr +oo)[f(x)-mx]#
#m=lim_(x rarr +oo)f(x)/x=lim_(x rarr +oo)log(x+2)/x=(+oo)/(+oo)#
Applying The L'Hopital's rule
#lim_(x rarr +oo)(h(x))/(i(x))=lim_(x rarr +oo)(h'(x))/(i'(x))#
#lim_(x rarr +oo)log(x+2)/x=lim_(x rarr +oo)(1/(x+2))/1=#
#m=lim_(x rarr +oo)1/(x+2)=0#
#q=lim_(x rarr +oo)[f(x)-mx]=lim_(x rarr +oo)[log(x+2)-0x]=#
#=lim_(x rarr +oo)log(x+2)=+oo#
It is not finite, then #cancel(EE)# slant asymptote
