limx->0 (sin3x.sin5x)/(x^2.cos2x)

Question

limx->0 (sin3x.sin5x)/(x^2.cos2x)

Answer 1

Recall that lim_θ→0[sinθ/θ] = 1

 

lim_x→0[sin(3x)/x]lim_x→0[sin(5x)/x]lim_x→0[1/cos(2x)]

 

= (3lim_x→0[sin(3x)/(3x)])(5lim_x→0[sin(5x)/(5x)])(lim_x→0[1/cos(2x)])

 

= (3)(1)(5)(1)(1) = 15