We have to find the value of lim x-->0 [(sin 2x - sin 6x) / 4x]

We see that substituting x = 0 at this stage gives an indeterminate form 0/0.

So we use l'Hopital's rule and substitute the numerator and the denominator with their derivatives.

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

substituting x = 0 here, we get (2 - 6)/4 = -1

**The required value of lim x-->0 [(sin 2x - sin 6x) / 4x] = -1**

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**