cara dan jawbannya gimana
Matematika
Priyo1111
Pertanyaan
cara dan jawbannya gimana
1 Jawaban
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1. Jawaban whongaliem
[tex] \lim_{x \to \ \frac{ \pi }{2} } \frac{4 sin^{2} x - 3.sin x - 1}{3.cos x . sin 2 x} = \lim_{x \to \ \frac{ \pi }{2} } \frac{(4sin x + 1)(sin x - 1)}{3.cos x .2.sin x .cos x} [/tex]
[tex]= \lim_{x \to \ \frac{ \pi }{2} } \frac{(4.sin x + 1)(sin x - 1)}{6 cos^{2} x . sin x} [/tex]
[tex]= \lim_{x \to \ \frac{ \pi }{2} } \frac{(4.sin x + 1)(sin x - 1)}{6 (1 - sin^{2} x) .sin x } [/tex]
[tex]= \lim_{x \to \ \frac{ \pi }{2} } \frac{(4.sin x + 1)(sin x - 1)}{- 6( sin^{2} x - 1) .sin x} [/tex]
[tex]= \lim_{x \to \ \frac{ \pi }{2} } \frac{(4.sin x + 1)(sin x - 1)}{- 6(sin x + 1)(sin x - 1).sin x} [/tex]
[tex]= \lim_{x \to \ \frac{ \pi }{2} } \frac{4.sin x + 1}{ - 6(sin x + 1) . sinx} [/tex]
[tex]= \frac{4.sin \frac{ \pi }{2} + 1}{- 6(sin \frac{ \pi }{2} + 1) . sin x} [/tex]
[tex]= \frac{4.1 + 1}{- 6( 1 + 1) .1} [/tex]
[tex]= \frac{4 + 1}{- 6 . 2 . 1} [/tex]
[tex]= - \frac{5}{12} [/tex]