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
