Tolonggg caranyaaa .

limit trigonometri
10. x -> 0  = 1/(4+3) = 1/7
11. x-> ~ = 2x/4x = 1/2
dilampiran 1 dan 2

sudut relasi :
cos (90° - x) = sin x
cos (-x) = cos x
cos (x - 90°) = cos (-(90° - x)) = cos (90° - x) = sin x

10) Lim (cos (x - 90°) cos x) / (4x + 3x cos x)
.... x => 0

Lim (sin x cos x) / x(4 + 3 cos x)
x => 0

Lim (sin x)/x . (cos x)/(4 + 3 cos x)
x=>0

= 1 . (cos 0)/(4 + 3 cos 0)
= 1 . 1/(4 + 3.1)
= 1/7

11) 1 - cos ax = 2 sin^2 1/2 ax

Lim x (1 - cos 1/√x)
x=>~

Lim 1/(1/x) . (2 sin^2 (1/2 . 1/√x)) ===========> 1/x = 1/√x . 1/√x
x=>~

Lim (2 sin (1/2 . 1/√x) . sin (1/2 . 1/√x)) / (1/√x . 1/√x)
x=>~

Lim 2 . [sin 1/2 (1/√x)] / (1/√x) . [sin 1/2 (1/√x)] / (1/√x)
x=>~

= 2 . 1/2 . 1/2

= 1/2