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