Question

Mohon bantuannya kakak:)​

Answer 1

Jawab:

limit trigonometri

Penjelasan dengan langkah-langkah:

a. lim (x → π/3)  ( sin x + 2 cos x ) / ( 3 tan x)

substitusikan x= π/3

limit =  ( sin π/3 + 2 cosπ/3 ) / ( 3 tan π/3)

limit =  (1/2 √3 + 2 (1/2)) /  (3 ( 1/3 √3))

limit =  (1/2 √3  + 1 ) / (√3)

limit =  { √3 ( 1/2 √3   + 1) / (3)

limit =  ( 3/2  +√3  ) /3

limit = 1/6 ( 3 + √3)

.

b. lim (x→0)  ( 1 - cos³ x ) / (sin² x)

lim (x→0)  {( 1 - cos x) (1+ cos x + cos² x)} / (1 - cos² x)

lim (x→0)  {( 1 - cos x) (1+ cos x + cos² x)} / (1 - cos x)(1 + cos x)

lim (x→0)  {(1+ cos x + cos² x)} / (1 + cos x)

x= 0  , limit = ( 1 + cos 0 + cos² 0 ) / ( 1 + cos 0)

limit = (1 + 1 + 1²)/ (1 + 1)

limit = 3/2

.

c)  lim (x→ 0) { 3x sin² 6x } / (tan³ 2x)

= (3x. (6x)² } / (2x)³

= (108 x³)  / (8 x³)

= 13,5