Matematika

Pertanyaan

1. Lim x=0 (x2-4) tan 3x / x3+5x2+6x 2. Lim x=0 cos 4x-cos 6x / cos 2x.sin2 5x

1 Jawaban

  • limit trigonometri

    1. lim(x->0)  { (x²-4) . tan 3x ] / (x³ + 5x² + 6x)
    = lim(x->0) { (x+2)(x-2). tan 3x " / (x)(x+2)(x+3)
    = lim (x->0)  (x-2) . tan 3x / (x+3). (x)
    = lim (x->0)  (x-2)/(x+3) .  lim(x->0) tan 3x/ x
    x= 0 --> (-2/3) . (3x/x) = - 2/3 .(3)
    limit = -2

    2. lim(x->0) {{ cos 4x - cos 6x)} / ( cos 2x. sin² 5x)
    = lim(x->0) - {cos 6x - cos 4x) / (cos 2x. sin² 5x)
    = lim(x->0) - (- 2 sin 5x. sin x)  (cos 2x . sin 5x. sin 5x)
    = lim(x->0  2/cos 2x . lim (x->0) (sin 5x. sin x)/(sin5x. sin 5x)
    x = 0
    = (2/ cos 0) . (5x. x)/(5x.5x)
    = (2/1).(1/5)
    = 2/5

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