Pertanyaan
tentukan limit dari. 1. lim x mendekati π cos2x-1 / sin x 2. lim x mendekati π/3 x+π/3 / tan (x+π/3).
3. lim x mendekati 0 sin x / sin 2x 4. lim x mendekati 0 1-cos 2x / sinx
5. lim u mendekati π/2 tan u / u
3. lim x mendekati 0 sin x / sin 2x 4. lim x mendekati 0 1-cos 2x / sinx
5. lim u mendekati π/2 tan u / u
Ditanyakan oleh: USER6786
248 Dilihat
248 Jawaban
Jawaban (248)
LImit trigonometri
1.limx->π) (cos 2x -1)/sin x
lim(x-> π) (-2 sin² x)/ (sin x)
= lim(x->π) -2 sin x
x= π --> limit = -2 sin π = - 2 (0)= 0
2.limit(x->π/3) (x+π/3)/ tan (x+π/3)
= (x+π/3)/ (x+π/3)
= 1
3. limit (x->0) sin x /sin 2x
= x/2x
= 1/2
4. lim(x->0) (1- cos 2x)/(sin x)
= lim(x->0) (2 sin² x)/(sin x)
= lim(x->0) 2 sin x
x =0 --> 2 sin 0 = 2(0)+ 0
5. lim(u-> pi/2) tan u/ u
= u/u = 1
1.limx->π) (cos 2x -1)/sin x
lim(x-> π) (-2 sin² x)/ (sin x)
= lim(x->π) -2 sin x
x= π --> limit = -2 sin π = - 2 (0)= 0
2.limit(x->π/3) (x+π/3)/ tan (x+π/3)
= (x+π/3)/ (x+π/3)
= 1
3. limit (x->0) sin x /sin 2x
= x/2x
= 1/2
4. lim(x->0) (1- cos 2x)/(sin x)
= lim(x->0) (2 sin² x)/(sin x)
= lim(x->0) 2 sin x
x =0 --> 2 sin 0 = 2(0)+ 0
5. lim(u-> pi/2) tan u/ u
= u/u = 1