~~Limit Trigonometri~~.... Question from @ZesshiZetsumei

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Jawaban:

Hasil limit = 7

Penjelasan dengan langkah-langkah:

Rumus dasar

$\lim _{x \to0} \frac{ax}{ \sin bx} = \lim _{x \to0} \frac{ \sin ax}{bx} = \frac{a}{b} \\ \lim _{x \to0} \frac{ax}{ \tan bx} = \lim _{x \to0} \frac{ \tan ax}{bx } = \frac{a}{b}$

Penyelesaian

$\lim _{x \to0} \frac{ \sin4x. { \tan}^{2}3x + 6 {x}^{3} }{2 {x}^{2} . \sin3x. \cos x} \\ = \lim _{x \to0} \frac{ \sin4x. { \tan}^{2}3x }{2 {x}^{2} . \sin3x. \cos x} + \lim _{x \to0} \frac{6 {x}^{3} }{2 {x}^{2}. \sin3x. \cos x } \\ = ( \lim _{x \to0} \frac{ \sin4x}{2x} \times \frac{ \tan3x}{x} \times \frac{ \tan3x}{ \sin3x} \times \frac{1}{ \cos x} ) + ( \lim _{x \to0} \frac{6x}{2x} \times \frac{x}{x} \times \frac{x}{ \sin3x} \times \frac{1}{ \cos x} ) \\ = ( \frac{4}{2} \times \frac{3}{1} \times \frac{3}{3} \times \frac{1}{ \cos0} ) + ( \frac{6}{2} \times \frac{1}{1} \times \frac{1}{3} \times \frac{1}{ \cos0} ) \\ = 6 + 1 = \textbf{7}$

_______________________

Detil jawaban

Mapel : Matematika
Kelas : 11
Materi : Bab 8 - Limit Fungsi Aljabar
Kata kunci : Limit trigonometri
Kode soal : 2
Kode kategorisasi : 11.2.8

11 votes Thanks 13

ZesshiZetsumei Terima Kasih kak ^_^

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aiya77 Halo guys

DB45

Jawab:

limit trigo bentuktak hingga

Penjelasan dengan langkah-langkah:

lim(x->0) { sin 4x tan² (3x) + 6 x³ } / { 2x² sin 3x cos x )

= lim(x->0) { sin 4x tan² (3x) + 6 x³ } / { 2x² sin 3x } X lim(x->0) 1/( cos x)

= lim(x->0) { sin 4x tan² (3x) + 6 x³ } / { 2x² sin 3x ) X 1

= ( 4x. (3x)² + 6x³ ) / (2x² )(3x)

= (36x³ + 6x³)/ (6x³)

= 42x³/6x³

= 7

5 votes Thanks 9

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Detil jawaban

Asal usul rumus limit tak hingga : lim (x→∞) ³√(ax³ + bx² + cx + d) - ³√(ax³ + px² + qx + r) = (b - p)/(3 ³√a²)

~Limit Fungsi Tak Hingga~ NO 36 pakai cara ya :)

~Limit Fungsi Trigonometri~ pakai permisalan yaa, jangan pakai turunan :)

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