If this is in Calculus II, you can use L'Hopital's Rule. If this is Calculus I, it will take a while.
The answer will be 0. It is really easy if you use L'Hopital's Rule: d/dx for both top and bottom.
(1-cos2x-cos4x) becomes (2sin2x + 4sin4x) and x becomes 1. As you put 0 for x, it will approach 0.
Remember, you can find special trigonometric limits from your textbook (if you are in Calc I):
lim (sin x)/x = 1 lim (1 - cos x)/x = 0 lim (1) / (cos x) = 1
x->0 x->0 x->0
Now, you can break / separate them into two:
lim (1-cos2x) - (cos4x)
x->0 x x
As you can see, first one approaches 0 (by looking at second limit above). Second one should approach 0 as well.
You have to use a lot of trig identities to solve this.