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14. SOAL-SOAL LIMIT FUNGSI                               UAN2006                                                          ...
Lim ( x 2 − x − 6)(4 + 5 x + 1)                                    Lim        tan 2 x Lim sin 4 x Lim sin 4 x=            ...
arahkan menjadi persamaan:                                                   UAN2006                                      ...
π                                                     F  (x) = 1. sin(2x-6) +(x-7) cos(2x-6). 2Untuk x =          - 2t dan...
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14. Soal-soal Limit Fungsi

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14. Soal-soal Limit Fungsi

  1. 1. 14. SOAL-SOAL LIMIT FUNGSI UAN2006 Lim x + 5 − 2x + 1 3. = ….EBTANAS2000 x→4 x−4 1 1 1 1 Lim x 2 − 5x + 6 A. - B. - C. 0 D. E.1. = …. 6 12 12 6 x → 2 x 2 − 3x + 2 jawab: 1 A. -1 B. - C. 0 D. 1 E. -5 3 Lim x + 5 − 2x + 1 jawab: x→4 x−4 0 kalau dimasukkan nilai x=2 didapat hasil = 0 Gunakan L’HOSPITAL Lim x + 5 − 2x + 1 x + 5 + 2x + 1 = . x→4 x−4 x + 5 + 2x + 1 Lim x − 5 x + 6 2 Lim 2 x − 5 = Lim x + 5 − (2 x + 1) x → 2 x − 3x + 2 x → 2 2 x − 3 2 = x → 4 ( x − 4)( x + 5 + 2 x + 1) 2.2 − 5 − 1 = = = -1 2.2 − 3 1 Lim −x+4 = x → 4 ( x − 4)( x + 5 + 2 x + 1) jawabannya adalah AUMPTN2000 Lim − ( x − 4) = x 2 − 2x Lim x → 4 ( x − 4)( x + 5 + 2 x + 1)2. Jika f(x) = , maka f(x) = …. x −4 2 x→2 Lim −1 = 1 x → 4 ( x + 5 + 2 x + 1) A. 0 B. ~ C. -2 D. E. 2 2 −1 1 1 = =- =- jawab: ( 4 + 5 + 2 .4 + 1 ) 3+3 6 Cara 1 : dengan L’HOSPITAL UN2007 Lim x2 − x − 6 Lim x 2 − 2 x Lim 2 x − 2 4. Nilai = …. = x → 3 4 − 5x + 1 x→2 x −42 x → 2 2x = 2.2 − 2 = 1 A. -8 B. -6 C. 6 D. 8 E. ~ 2.2 2 Cara 2 : pemfaktoran jawab: Lim x 2 − 2 x Lim x( x − 2) Lim x 2 − x − 6 = x→2 x −42 x → 2 ( x − 2)( x + 2) x → 3 4 − 5x + 1 Lim x Lim x 2 − x − 6 4 + 5 x + 1 = = x → 2 ( x + 2) x → 3 4 − 5x + 1 4 + 5x + 1 x 2 1 = = = ( x + 2) (2 + 2) 2 Lim ( x 2 − x − 6)(4 + 5 x + 1) = x→3 16 − (5 x + 1)jawabannya adalah D www.matematika-sma.com - 1
  2. 2. Lim ( x 2 − x − 6)(4 + 5 x + 1) Lim tan 2 x Lim sin 4 x Lim sin 4 x= = -2 . . x→3 16 − (5 x + 1) x→0 16 x x → 0 x x→0 x Lim ( x − 3)( x + 2)(4 + 5 x + 1) 2= = -2 . . 4 . 4 = -4 x→3 15 − 5 x 16 jawabannya adalah A Lim ( x − 3)( x + 2)(4 + 5 x + 1)= x→3 − 5( x − 3) UN2002 Lim 1 − cos 2 ( x − 2) 6. =… Lim ( x + 2)(4 + 5 x + 1) x → 2 3x 2 − 12 x + 12= x→3 −5 1 1 A. 0 B. C. 3 D. 1 E. 3 (3 + 2)(4 + 5.3 + 1) 3 3= −5 5(4 + 4) 40 jawab:= = = -8 −5 −5 cos 2 x + sin 2 x =1 ⇔ cos 2 (x-2) + sin 2 (x-2) = 1UAN2005 Lim tan 2 x cos 8 x − tan 2 x ⇒ cos 2 (x-2) = 1 - sin 2 (x-2)5. Nilai dari =… x→0 16 x 3 Lim 1 − cos 2 ( x − 2) A. -4 B.-6 C.-8 D.-16 E.-32 x → 2 3x 2 − 12 x + 12 jawab: Lim 1 − (1 − sin 2 ( x − 2)) = Lim tan 2 x cos 8 x − tan 2 x x → 2 3x 2 − 12 x + 12 x→0 16 x 3 Lim 1 − 1 + sin 2 ( x − 2) = Limtan 2 x(cos 8 x − 1) x → 2 3 x 2 − 12 x + 12 = x→0 16 x 3 Lim sin 2 ( x − 2) = cos 8x =cos(4x+4x) x → 2 3( x 2 − 4 x + 4) = cos 4x . cos 4x – sin 4x . sin4x = cos 2 4x - sin 2 4x Lim 1 sin 2 ( x − 2) 1 = 1 - sin 2 4x - sin 2 4x = = x → 2 3 ( x − 2) 2 3 = 1 - 2 sin 2 4x Jawabannya adalah B Lim tan 2 x(cos 8 x − 1) UAN2005 x→0 16 x 3 Lim 7. Nilai {(3x-1) - 9 x 2 − 11x + 9 }= … x →~ Lim tan 2 x(1 − 2 sin 2 4 x − 1) = x→0 16 x 3 1 3 5 A. -1 B. 0 C. D. E. 6 6 6 Lim tan 2 x(−2 sin 2 4 x) = x→0 16 x 3 Jawab: www.matematika-sma.com - 2
  3. 3. arahkan menjadi persamaan: UAN2006 π cos x − sin Limx →~ ( ax 2 + bx + c − ax 2 + px + q = ) b− p 2 a 9. Nilai Lim x→ π π x 6 = …. 3 − 6 2 Lim 1 1 {(3x-1) - 9 x 2 − 11x + 9 } A. 3 B. 3 C. 3 D.-2 3 E. -3 3x →~ 2 3 Lim jawab:= { (3x − 1) 2 - 9 x 2 − 11x + 9 } x →~ 0 Kalau nilai x dimasukkan didapat nilai: Lim 0= { 9x 2 − 6x + 1 - 9 x 2 − 11x + 9 } x →~ Cara 1: L’Hospital b− p − 6 − (−11) 5 π= = = Lim cos x − sin 2 a 2 9 6 π 6 x→ π x 3 −Jawabannya adalah E 6 2EBTANAS1994 Lim Lim − sin x Lim 3x − 5 = π = π 2 sin x8. =…. x→ 1 x→ x →~ 2 x + 4 x + 5 2 3 − 3 2 8 3 A. 0 B. C. D. 1 E. 6 π 1 11 4 = 2 . sin = 2. 3 = 3 3 2 jawab: Cara 2: pemfaktoran (agak panjang) dibahas disini sebagai perbandingan: rumus dasar: π x Lim ax m + bx m −1 + ... Dimisalkan : − =t 6 2 x →~ px n + qx n −1 + ... x π maka : = -t 2 6 membagi pembilang dan penyebut dengan pangkat π tertinggi penyebut x=2( - t) 6 3x 5 π − = - 2t Lim 3x − 5 Lim x2 x2 3 = x → ~ 2 x 2 + 4 x + 5 x →~ 2 x 2 4 x 5 = π + 2 + 2 π π π π x2 x x untuk nilai x = maka t = − 3 = − =0 3 5 3 6 2 6 6 Lim − = x x2 x →~ 4 5 2+ + 2 x x 0−0 0 = = =0 2+0+0 2Jawabannya adalah A www.matematika-sma.com - 3
  4. 4. π F (x) = 1. sin(2x-6) +(x-7) cos(2x-6). 2Untuk x = - 2t dan t → 0 , maka 3 G (x) = 2x + 2 π Lim cos x − sin Lim F ( x) Lim sin(2 x − 6) + 2( x − 7) cos(2 x − 6) π 6 =x→ π x x → 3 G ( x) x→3 2x + 2 3 − 6 2 π 0 + 2(−4).1 − 8 Lim cos( 3 − 2t ) − 2 1 = = = -1= 2.3 + 2 8 t→0 t π π Lim cos 3 cos 2t + sin 3 sin 2t − 2 1= t→0 t 1 1 Lim 2 cos 2t + 3 sin 2t − 1 2 2= t→0 t 1 1 Lim 2 (1 − 2 sin t ) + 2 3 (2 sin t cos t ) − 2 2 1= t→0 t 1 Lim 2 − sin t + 3 sin t cos t − 2 2 1= t→0 t Lim − sin 2 t + 3 sin t cos t= t→0 t Lim sin t (− sin t + 3 cos t )= t→0 t Lim sin t Lim= . (-sin t + 3 cos t) t→0 t t→0= 1 . (0 + 3)= 3Jawabannya adalah CUAN2004 Lim ( x − 7) sin(2 x − 6)10. Nilai = …. x→3 x 2 + 2 x − 15 A. -4 B. -1 C. 0 D. 1 E.4 jawab: cara yang cepat dengan menggunakan L’Hospital dengan catatan kita harus menguasai differensial/turunan Cara ini cocok untuk soal multiple choice seperti ini. Lim F ( x) x → 3 G ( x) Ingat : y = uv, maka y = u v + u v www.matematika-sma.com - 4

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