nilai daru limx->0 ( 1-cos4x) sin x/x² tan 2x
Matematika
aailmi1
Pertanyaan
nilai daru limx->0 ( 1-cos4x) sin x/x² tan 2x
1 Jawaban
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1. Jawaban whongaliem
limit (1 - cos 4x) .sin x / x² .tan 2x = limit {1 - (cos² 2x - sin² 2x)} sin x / x².tan 2x
x⇒0 x⇒0
= limit (1 - cos² 2x + sin²2x) .sin x / x² .tan 2x
x⇒0
= limit (sin² 2x + sin² 2x).sin x / x² . tan 2x
x⇒0
= limit 2.sin² 2x . sin x / x² .tan 2x
x⇒0
= limit 2.sin² 2x / x² .sin x / x . x / tan 2x
x⇒0
= 2 . 2² . 1 . 1/2
= 2²
= 4