Skorr

Limit trigonometri

Kalkulus Matematika · SMA hard

Nilai dari limx3x2+6x+922cos(2x+6)\displaystyle\lim_{x\to-3} \dfrac{x^2+6x+9}{2-2\cos(2x+6)} adalah ....

  1. A
    3
  2. B
    1
  3. C
    12\dfrac12
  4. D
    13\dfrac13
  5. E
    14\dfrac14

Kunci jawaban

E

Penjelasan singkat

x2+6x+9=(x+3)2x^2+6x+9=(x+3)^2; 22cos(2(x+3))=4sin2(x+3)2-2\cos(2(x+3))=4\sin^2(x+3). Limit =(x+3)24(x+3)2=14=\dfrac{(x+3)^2}{4(x+3)^2}=\tfrac14.

(Jawaban dihitung & diverifikasi dengan Python/SymPy.)

Pembilang =(x+3)2=(x+3)^2. Penyebut =2(1cos(2(x+3)))=22sin2(x+3)=4sin2(x+3)=2\bigl(1-\cos(2(x+3))\bigr)=2\cdot2\sin^2(x+3)=4\sin^2(x+3).

limx3(x+3)24sin2(x+3)=14limu0(usinu)2=14\lim_{x\to-3}\dfrac{(x+3)^2}{4\sin^2(x+3)}=\dfrac14\lim_{u\to0}\left(\dfrac{u}{\sin u}\right)^2=\dfrac14

E.

⚡ Latihan Matematika SMA