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...

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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**