Question

Tolong bantu kawan kawan :)​

Answer 1

Jawab:

limit trigonometri bentuk tak tentu

Penjelasan dengan langkah-langkah:

lim(x-> π/2)  (tan 2x ) / ( sin 2x -  cos 3x )

misal t = x - π/2

i) x = π/2 + t

ii) x = π/2 --> t = 0

.

lim(x-> π/2)  (tan 2x ) / ( sin 2x -  cos 3x )=

= lim(t ->0)  [tan 2 (π/2 + t) } /  [ sin 2(π/2 + t) - cos 3(π/2 + t)]

= lim(t ->0)  [tan (π + 2t) } /  [ sin (π + 2t) - cos (3π/2 + 3t)]

= lim(t ->0)  [tan (2t) } /  [ - sin (2t) -sin (3t)]

= (2t)/ ( - 2t  - 3t)

=  2t /-5t

= -2/5