AP Calculus BC question

Question

find the limit using lim x->0 sinx/x = 1

lim x->0 (sin3xcot5x)/xcot4x

Answer 1

Note cot(5x) = cos(5x)/sin(5x) and 1/cot(4x) = sin(4x)/cos(4x). Thus

sin(3x) cot(5x)/(x cot(4x)) = sin(3x)/x * cos(5x)/cos(4x) * sin(4x)/sin(5x)

.........................................= 3 * sin(3x)/(3x) * cos(5x)/cos(4x) * 4 * sin(4x)/(4x) * (5x)/sin(5x) * (1/5)

which, in the limit as x -> 0, is 3 * 1 * 1/1 * 4 * 1 * 1 * 1/5 = 12/5.