# derivatives and limitevaluate this limit using derivatives (sinx-sin7x)/4x, x approaches to 0

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We have to find the value of lim x-->0 [ (sinx-sin7x)/4x]

Now if we were to substitute x = 0, we get the indeterminate form 0/0. So we can use l'Hopital's rule and substitute the numerator and denominator with their derivatives.

So we get lim x--> 0[ (cos x - 7*cos 7x)/4]

substitute x = 0

=> (1 - 7)/4 = -6/4 = -3/2

**The required limit is -3/2.**

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

lim (sinx-sin7x)/4x = (0 - 0)/0 = 0/0

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

lim (sinx-sin7x)/4x = lim (sinx-sin7x)'/(4x)'

lim (sinx-sin7x)'/4(x)' = lim (cos x- 7cos 7x)/4

We'll substitute x by accumulation point:

lim (cos x- 7cos 7x)/4 = (cos 0- 7cos 7*0)/4

lim (cos x- 7cos 7x)/4 = (1 - 7*1)/4

lim (cos x- 7cos 7x)/4 = -6/4** **

**For x->0, lim (sinx-sin7x)/4x = -3/2**