please tolong bantu dijawab kk

1) Lim (Tan 2x cos 8x - Tan 2x)/16x^2
= Lim Tan 2x (cos 8x - 1) / 16x^2
= Lim Tan 2x (-2 sin^2 4x) / 16x^2
= Lim (-2 Tan 2x sin 4x sin 4x) / 16x^2
= Lim -2/16 . (Tan 2x)/x . (Sin 4x)/x . Sin 4x
= -1/8 . 2/1 . 4/1 . Sin 0
= -1 . 0
= 0
Kemungkinan soal no 1 penyebutnya 16x^3 agar jawabannya ≠ 0
Lim (Tan 2x cos 8x - Tan 2x)/16x^3
= Lim Tan 2x (cos 8x - 1) / 16x^3
= Lim Tan 2x (-2 sin^2 4x) / 16x^3
= Lim (-2 Tan 2x sin 4x sin 4x) / 16x^3
= Lim -2/16 . (Tan 2x)/x . (Sin 4x)/x . (sin 4x)/x
= -1/8 . 2/1 . 4/1 . 4/1
= -4

2) no 2 jg kemungkinan ada kesalahan soal jg yaitu x => 5
Lim (4x - 10) sin (x - 5) / (x^2 - 25)
= Lim (4x - 10) sin (x - 5) / (x + 5)(x - 5)
= Lim (4x - 10)/(x + 5) . (sin (x - 5))/(x - 5)
= (4.5 - 10)/(5 + 5) . 1/1
= 10/10
= 1

Maaf ya ngerubah-rubah "soal"nya