Expert: Abe Mantell - 1/4/2010

Question
the question is
f(x)= ln(1-cosx)
...
I solved in this way....
...
this function is defined when (1-cosx)> 0 >>> -cosx>-1 ..>>
cosx<1 >>>>> x<0
so, the domain is (- infinity , 0)
x=0 is a vertical asymptote...

x ..> -infinity,, -1<= cosx <= 1 the value of cos(anything) is restricted between -1 and 1
by assuming that when x ..> -infinity ... cos(-infinity)= -1

x ..> -infinity, f(x) ..> ln(2)

This is what I am not convinced about at..... which f(-infinity)= ln(2)
..
so the horizontal line is y=ln2= 0.693......

Answer
Is your question: What is/are the asymptotes of the function f(x)=ln(1-cos(x))??

Yes, f(x) is defined as long as cos(x)<1 ==> x cannot be any multiple of 2pi, i.e.
x cannot be 2*pi*k, where k=any integer.  The function has infinitely many vertical
asymptotes: x=0, -2pi, +2pi, -4pi, +4pi, ...

The limit as x->infinity of f(x) does not exist, as 1-cos(x) oscillates between
0 and 2.

Abe