Expert: Sherry Wallin - 11/28/2008

Question
Lim   [sin(3x+a)-3sin(2x+a)+3sin(x+a)-sin(a)]/x^3
x->0

Answer
Hi Prashant~
    First you want to rewrite sin(3x   a) using the trig identity
sin (a b) = sin a cos b   cos a sin b and getting sin 3x cos a   cos 3x sin a and likewise with sin(2x a) and sin (x a). When all is said and done you will have:
lim{cos(a)[sin 3x-3sin 2x  3sinx] sin(a)[cos 3x-3cos 2x  3cosx -1]}/x^3
x->0
which if you rewrite the limit of a sum as the sum of the limits you can use L'Hopital's Rule on each term. Notice the first 3 terms have a constant of cos(a) times a function with sin(nx) in it, n = 1,2,3 and when you use L'Hopital's Rule you end up with:

(-9/2)cos(a)*lim cos 3x   4cos(a)*lim cos 2x -(1/2)cos(a)*lim cos x
            x->0                 x->0                    x->0
recalling that lim cos(nx) = 1
              x->0
you have (cos(a)[-9/2)*1  4*1 -(1/2)*1] = cos (a)(-5   4) = -1cos(a)
the other half of the original equation generates sin(nx) terms which goes to 0 as x goes to 0.

I hope this answers your question. If anything is unclear please ask me for help again.

Math Prof