Question

HELP, SOAL MTK MINAT ​

Answer 1

Jawab:

limit trigonometri bentu tak tentu

Penjelasan dengan langkah-langkah:

lim (x-> π/2) { 1 - sin x ] / (x - π/2 )

* ubah sin x = cos (π/2 - x)

=  lim (x-> π/2) { 1 - cos (π/2 - x) } / (x - π/2 )

= lim (x -> π/2)  [ 1 - { 1 - 2 sin² 1/2 (π/2 - x) }] / - (π/2 - x)

= lim (x -> π/2)   ( 2 sin² 1/2 (π/2 - x) ) / - (π/2 - x)

= lim(x-> π/2)  2 sin 1/2 (π/2 - x) sin 1/2 (π/2 - x) /  - (π/2 - x)

= 2 (1/2) . lim  (x ->0)  sin 1/2 (π/2 - x)

= 1 . sin 1/2 (0)

= 1 (0)

= 0