Determine the limit of the function (sin5x-sin3x)/x, x-->0

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

lim (sin5x-sin3x)/x = (0 - 0)/0 =
0/0

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

lim (sin5x-sin3x)/x = lim
(sin5x-sin3x)'/(x)'

 lim (sin5x-sin3x)'/(x)' = lim (5cos
5x- 3cos 3x)/1

We'll substitute x by accumulation
point:

lim (5cos 5x- 3cos 3x) = 5cos5*0 - 3cos
3*0

lim (5cos 5x- 3cos 3x) = 5*1 -
3*1

lim (5cos 5x- 3cos 3x) =
2

For x->0, lim (sin5x-sin3x)/x =
2.