Page 13 - EbooK MATEMATIKA PEMINATAN KELAS XII Yovy shelviani

2 2
1− cos 2x = lim 1− (1− 2 sin x) ................... sin x
lim = lim = lim

2
2
x→0 1− cos 4x x→0 1− (1− 2 sin 2x) x→0 ................... x→0 (2 sin x cos x)
• 1 1
= lim ................. = lim =
2
x→0 ............................ x→0 4 cos x 4
 Sudut rangkap
sin(  − x) sin[−(x − 2 )]
• lim cos x = lim 2 = lim = 1
      2 2
x→ − x x→ − x x→ 2 [−(x − )] cos 2a = cos a ‒ sin a
2 2 cos 2a = 2 cos a ‒ 1
2
2 2 2
2
cos 2a = 1 ‒ 2 sin a
1 1 2 tan a
sin(x − 2) = lim 1 sin(x − 2) .(1) = tan2a = 2
• lim = 1−tan a

x→2 x 2 x→2 x + 2 x − 2 2 + 2 4

Uji Kompetensi 1.1
Sederhanakanlah dan selesaikanlah limit-limit dibawah ini:
sin 6x tan 2x
1) lim = 8) lim =
x→0 2x x→0 5x
tan 4x tan 2x
2) lim = 9) lim = ...
x→0 3x x→0 5x
2
tan 2x.tan3x sin 2x
3) lim = ... 10) lim 2 = ...
x→0 x.tan x x→0 tan 3x

sin(x − )

4) lim 4 = ... 11) lim (3x +1).sin(x −1) = ...

  2
x→ (x − ) x→1 x + 2x − 3
4
4

sin(x + ) 
3
5) lim  = ... 12) lim cos(x − ) = ...


x→− (x + ) x→ 3
3 2
3 
2
6) lim sin(2x − ) = ... 13) lim sin (x − ) =
 
2
x→ x→ 4
2
2
(x −1) sin 6x x − sin(x − 3) − 3
7) lim 3  = ... 14) lim ) =
x→0 x + 3x + 2 x→3 x − 3
Uji Kompetensi 1.2

Sederhanakanlah dan selesaikanlah limit-limit dibawah ini: Kesamaan setengah sudut
1 sin( ) =  1 − cos x
x
sin x tan 2 x 2 2
1. lim 2 = ...(1) x 1 + cos x
x→0 x x cos( ) =  2
2

2 n
(cos 2x −1) cos nx = 1− 2sin ( x), nR
2
2. lim = ... sifat identitas [‒ 2 sin a] 2
x→0 x 2

11 | Matematika Peminatan SMA/MA Kelas XII