contoh limit trigono metri
TRANSCRIPT
Jl.Pekayon Raya / Perumahan Pondok Suryamandala Blok Q1 No 24Kel.Jakamulya Bekasi Selatan ( 021) 8209381
CONTOH SOALLIMIT FUNGSI TRIGONOMETRI
1.
lim
x→ π2
1−sin2 x(sin 1
2x−cos 1
2x )2
=. .. .. .
Jawab :
lim
x→ π2
(1−sin x )(1+sin x )sin2 1
2x+cos2 1
2x−2sin 1
2xcos 1
2x=lim
x→ π2
(1−sin x )(1+sin x )1−sin x
=1+1=2
2.
limx→0
sin 6 xsin 2 x
=.. .. .
Jawab :
limx→0
sin 6 xsin 2 x
=62=3
3.
limx→0
cos 2x−1x2
=. .. .. .
Jawab :
limx→0
1−2sin2 x−1x2
=limx→0
−2 . sin xx.sin xx
=−2 .1 .1=−2
4.limx→2
sin( x−2)x2−4
=. . .. .. .
Jawab :
limx→2
sin( x−2)( x−2 )(x+2 )
=limx→2
sin( x−2 )x−2
.1x+2
=1 . 12+2
=14
5.
lim
x→ π4
cos2 xsin x−cos x
= .. .. .. . ..
Jawab :
limx→ π
4
(cos x−sin x )(cos x+sin x )−(cos x−sin x )
=12√2+ 1
2√2
−1=−√2
6.limx→0
sin axsin bx
=. .. .. ..
Jawab : limx→0
sin axsin bx
.bxax.ab=limx→0
sinaxax
.bxsin bx
.ab=1.1. a
b=ab
7.
limx→0
sin 2 x3−√2 x+9
=. .. .. . .
Jawab :
limx→0
2sin x cos x3−√2 x+9
.3+√2 x+93+√2 x+9
=limx→0
sin xx
.cos x .(3+√2 x+9)−1
=1.1. (3+3)−1
=−6
8.limx→0
1−cos xx sin 2 x
=. .. . .. ..
Jawab :
limx→0
2sin2 12x
x sin 2x=limx→0
sin 12x
12x.sin 1
2x
sin2 x=1.
12
2=14
9.
limx→0
x tan x1−cos 2x
=. .. .. . ..
Jawab : limx→0
x tan x
2sin2 x=limx→0
12.xsin x
.tan xsin x
=12
10.
limx→0
tan x
x2+2 x=. . .. .. .
Jawab :
limx→0
tan xx.1x+2
=1 . 12=12
11.
limx→0
1−cos x5x2
=. .. . ..
Jawab :
limx→0
2sin2 12x
5 x2=25.12.12= 110
12.
limx→0
sin x
√1−x−1=. .. .. .
Jawab :
limx→0
sin x√1−x−1
. √1−x+1√1−x+1
=limx→0
sin xx. √1−x+1
−1=1 . 1+1
−1=−2
13.
limx→0
cot xcot 2 x
=. . .. ..
Jawab :
limx→0
1tan x1
tan2 x
=limx→0
tan 2 xtan x
=2
14.
limx→0
sin 4 x+sin 2x3 x cos x
=.. .. . .
Jawab :
limx→0
2sin 3x cos x3 x cos x
=2.1=2
15.limx→0
x sin x1−cos 4 x
=. .. .. . .
Jawab : limx→0
x sin x
2sin22 x=limx→0
12.x
sin 2 x.sin xsin 2x
=12.12.12=18
16.limx→0
cos 4 x−1x tan 2 x
=. .. .. . .
Jawab :
limx→0
−2sin22 xx tan2 x
=limx→0
−2. sin 2xx
.sin 2xtan 2 x
=−2 . 21.22=−4
17.
limx→0
sin22xx2cos2 x
=. . .. .. . .
Jawab : limx→0
sin 2 xx
.sin 2 xx
.1
cos2 x=21.21.11=4
18.limx→0
7 x2+sin(2 x2 )tan23 x
=. . .. .. . .
Jawab :
limx→0
(√7 x )2( tan 3x )2
+sin(√2x )2( tan 3 x )2
=79+ 29=1
19.
limx→0
cos 4 x−1cos 5x−cos3 x
=. .. .. . ..
Jawab :
limx→0
−2sin22x−2sin 4 x sin x
=limx→0
sin 2xsin 4 x
.sin2 xsin x
= 24.21=1
20.
limx→0
4 xx+sin 3x
=. .. . .. ..
Jawab : limx→0
1x+sin 3 x4 x
=limx→0
114+ sin 3 x
4 x
= 114+ 34
=1
21.limx→0
sin (2x2 )x2+sin23 x
=. .. .. . ..
Jawab : limx→0
1
x2
sin 2x2 +sin
23 x
sin (√2 x )2= 112+92
=15
22.
limx→0
sin 4 x . tan23x+6 x3
2x2 . sin 3 x . cos2 x=. .. . .. .. ..
Jawab :
limx→0
sin 4 xsin 3 x
.tan23 x(√2x )2
.1cos2x
+62.x2
x2.xsin 3x
.1cos2 x
¿43.92.1+3 .1.1
3.1=7
23.limx→0
1−cos2 x−cos x sin2 xx4
=. .. .. . ..
Jawab :
limx→0
sin2 x (1−cos x )x 4
=limx→0
2 .sin2 xx2
.sin2 1
2x
x2=2 .12 .( 1
2)2=1
2
24.limx→1
sin (1− 1x)cos (1− 1
x)
x−1=. .. .. .. .
Jawab :
limx→1
sin (1− 1x)cos (1− 1
x)
x (1− 1x)
=limx→1
sin (1− 1x)
1− 1x
.cos (1− 1
x)
x=1.1=1
25.
limx→1
sin (πx−π )( x−1)cos ( πx−π )
=. .. .. .. .
Jawab :
limx→1
sin π ( x−1 )x−1
.1
cos (πx−π )=π .1=π