Question

Limit x tends to 0 . Cos4x - cos3x / x ?

Answer 1

Assuming that it should be (cos4x-cos3x)/x, please see below.

Explanation:

`cos4x = cos(3x+x) = cos(3x)cos(x)-sin(3x)sin(x)`, so we have

`(cos4x-cos3x)/x = (cos(3x)cos(x)-sin(3x)sin(x)-cos(3x))/x`

`= (cos(3x)cos(x)-cos(x))/x - (sin(3x)sin(x))/x`

`= 3cosx(cos3x-1)/(3x) - sin(3x)(sinx/x)`

The limit at `0` is

`3(1)(0)-(0)(1) = 0`

Otherwise

the limit as `xrarr0` of cos4x-cos3x/x does not exist. It is `+-oo` for the left and right limits.

Answer 2

I assume you mean:

`lim_(x->0) (cos 4x - cos3x)/x`

The limit is in the indeterminate form `0/0` so we can use l'Hospital's rule:

`lim_(x->0) (cos 4x - cos3x)/x = lim_(x->0) (d/dx(cos 4x - cos3x))/(d/dx x) = lim_(x->0) (-4sin4x +3sin3x) = 0`

Alternatively we can use Taylor's expansion:

`lim_(x->0) (cos 4x - cos3x)/x = lim_(x->0) (1-8x^2+o(x^2) -1 +9/2x^2 +o(x^2))/x`

`lim_(x->0) (cos 4x - cos3x)/x = lim_(x->0) -(7x)/2 +1/x o(x^2) = 0`