limx->0 (cosx-cos11x)/(cos3x-cos7x)

Question

limx->0 (cosx-cos11x)/(cos3x-cos7x)

Answer 1

Limit as x→0 of (cosx - cos11x) / (cos3x - cos7x) = ?

 

This is an indeterminate form, so use L'Hopital's rule.

 

(cosx - cos11x) / (cos3x - cos7x) → (-sinx + 11sin11x) / (-3sin3x + 7sin7x)

 

Note the form is still indeterminate.  Apply L'Hopital's rule again.

 

(-sinx + 11sin11x) / (-3sin3x + 7sin7x) → (-cosx + 121cos11x) / (-9cos3x + 49cos7x)

 

Now let x→0 in both numerator and denominator.  You get

 

(-1 + 121) / (-9 + 49) = 120/40 = 3

 

The limit is 3.