Question

Kak, tolong bantu dong menjawab soal nomor 2b dan 2c...
besok dikumpul nih kk

Answer 1

• Limit TriGono

Lim (cos 3x / 2x)
x->0

Lim (4cos³x - 3 cos x) / 2x
x->0

Lim [cos x (4cos² x - 3) / 2x]
x->0

Lim (1/2) . ( cos x / x) . (4 cos² x - 3)
x->0

= (1/2) . 1 . [ 4(1) - 3]

= 1/2


***

Lim ( tan 5x / 3x )
x->0

= 5/3

Catatan :
"Untuk Sin dan tan , bisa langsung dioperasikan"

Answer 2

Jawab:

limit trigonometri

lim(x->0) cos x = 1

lim(x ->0) (sin ax)/ bx =  a/b

lim(x->0) (tan ax)/bx = a/b

Penjelasan dengan langkah-langkah:

2b.

lim(x->0)  { cos 3x } / (2x)

= lim(x->0)  cos 3x  ( 1/2x)

x= 0 --> limit  = cos (0)  { 1 / (2.0)}

x= 0 --> limit =  1 ( 1/0)  = 1/0 = ∞

2c.

limit (x->0)   (tan 5x) /(3x) =

= lim (x ->0)  (tan 5x ) / (3x)  { 5x/5x}

=lim (x-> 0)  {tan 5x)/(5x) .  (5x)/(3x)

x= 0 --> limit = 1 (5/3) = 5/3