tolong bantuan secepatnya untuk belajar PAS
Matematika
ayunurdewi99
Pertanyaan
tolong bantuan secepatnya untuk belajar PAS
1 Jawaban
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1. Jawaban whongaliem
[tex] \lim_{x \to \ 0} \frac{1 - cos 4x}{4.cos x . sin^{2} x} = \lim_{x \to \ 0} \frac{1 - ( cos^{2} 2x - sin^{2} 2x)}{4.cos x sin^{2} x} [/tex]
= [tex] \lim_{x \to \ 0} \frac{1 - cos^{2} 2x + sin^{2} 2x}{4.cos x sin^{2} x} [/tex]
[tex]= \lim_{x \to \ 0} \frac{ sin^{2} 2x + sin^{2} 2x}{4.cos x sin^{2} x} [/tex]
[tex]= \lim_{x \to \ 0} \frac{2 sin^{2} 2x}{4.cos x . sin^{2} x} [/tex]
[tex]= \lim_{x \to \ 0} \frac{2 .( 2 .sin x .cos x)^{2} }{4.cos x . sin^{2} x} [/tex]
[tex]= \lim_{x \to \ 0} \frac{2 (4 . sin^{2} x . cos^{2} x)}{4.cos x . sin^{2} x} [/tex]
[tex]= \lim_{x \to \ 0} 2.cos x[/tex]
= 2 .cos 0°
= 2 . 1
= 2