Question

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

Accepted Answer

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

Answer

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

Calculus

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

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