# I have to solve the limit of the function (sinx-sin7x)/x x-->0 but i should not use derivatives and remarcable limits. How to solve the limit?

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### 1 Answer

First, we notice that if we'll substitute x by 0, we'll get an indetermination.

lim (sinx-sin7x)/x = (0 - 0)/0 = 0/0 (since sin 0 = 0)

Since we don't have to apply L'Hospital rule or remarkable limits, we'll transform the difference of functions from numerator, into a product, using the formula:

sin a - sin b = 2 cos [(a+b)/2]*sin [(a-b)/2]

sinx - sin7x = 2 cos [(x+7x)/2]*sin [(x-7x)/2]

sinx - sin7x = 2 cos [(8x)/2]*sin [(-6x)/2]

sinx - sin7x = -2 cos (4x)*sin (3x)

We'll re-write the limit:

lim (sinx-sin7x)/x = lim -2 cos (4x)*sin (3x)/x

-2lim cos (4x)*lim 3*sin (3x)/3x = -2*3lim cos (4x)*lim sin (3x)/3x

We'll substitute x by the value of accumulation point:

-2*3lim cos (4x)*lim sin (3x)/3x = -6*cos 0*1 = -6

**The value of the given limit, for x->0, is: lim (sinx-sin7x)/x = -6.**