What is the limit of the function (sin2x-sin6x)/4x, x-->0?

First, we'll verify if we'll have an indetermination by
substituting x by the value of the accumulation
point.

lim (sin2x-sin6x)/x = (0 - 0)/0 =
0/0

Since we've get an indetermination, we'll apply
l'Hospital rule:

lim (sin2x-sin6x)/4x = lim
(sin2x-sin6x)'/(4x)'

 lim (sin2x-sin6x)'/4(x)' = lim (2cos
2x- 6cos 6x)/4

We'll substitute x by accumulation
point:

lim (2cos 2x- 6cos 6x)/4 = (2cos 2*0- 6cos
6*0)/4

lim (2cos 2x- 6cos 6x)/4 = (2*1 -
6*1)/4

lim (2cos 2x- 6cos 6x)/4 =
-4/4

For x->0, lim (sin2x-sin6x)/4x =
-1