026.19. a) lim sin 3x + sin x x→ cos 3x + cos x π 2 ,; 6) lim cos 5x − cos 3x x→0 sin 5x + sin 3x

• а) lim (sin 3x + sin x) / (cos 3x + cos x), x → π/2 lim (sin (3π/2) + sin (π/2)) / (cos (3π/2) + cos (π/2)) = (-1 + 1) / (0 + 0) = 0/0 Применим правило Лопиталя: lim (3cos 3x + cos x) / (-3sin 3x - sin x) = (3cos (3π/2) + cos (π/2)) / (-3sin (3π/2) - sin (π/2)) = (0 + 0) / (3 - 1) = 0/2 = 0
• б) lim (cos 5x - cos 3x) / (sin 5x + sin 3x), x → 0 lim (cos 0 - cos 0) / (sin 0 + sin 0) = (1 - 1) / (0 + 0) = 0/0 Применим правило Лопиталя: lim (-5sin 5x + 3sin 3x) / (5cos 5x + 3cos 3x) = (-5sin 0 + 3sin 0) / (5cos 0 + 3cos 0) = (0 + 0) / (5 + 3) = 0/8 = 0

Ответ: a) 0; б) 0.