O slideshow foi denunciado.
Utilizamos seu perfil e dados de atividades no LinkedIn para personalizar e exibir anúncios mais relevantes. Altere suas preferências de anúncios quando desejar.

Tugas Calculus : Limit (Hal. 8-14)

Tugas Calculus : Limit (Hal. 8-14)

Livros relacionados

Gratuito durante 30 dias do Scribd

Ver tudo
  • Seja o primeiro a comentar

  • Seja a primeira pessoa a gostar disto

Tugas Calculus : Limit (Hal. 8-14)

  1. 1. TUGAS CALCULUS (HAL 8-14) D I S U S U N Oleh : Nama : Cinjy Saylendra Monica Dita Yoriza Indah Yanti Monica Roselina Prodi : Teknik Elektronika Kelas : 1E A Semester : 2 (dua) POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG Kawasan Industri Air Kantung Sungailiat, Bangka 33211 Telp. (0717) 93586, Fax. (0717) 93585 Email : polman@polman-babel.ac.id Website : www.polman-babel.ac.id TAHUN AJARAN 2014/2015
  2. 2. Latihan 2.2 1. lim 𝑥→∞ (5𝑥 − 7) = ∞ 2. lim 𝑥→∞ 7 𝑥3 = 7 ∞ = 0 3. lim 𝑥→∞ 3𝑥 + 95 = ∞ + 95 = ∞ 4. lim 𝑥→∞ 𝑥3−𝑥2+47𝑥+9 18𝑥3+76𝑥−11 = 1 18 5. lim 𝑥→∞ 8 4−𝑥 = 8 ∞ = 0 6. lim 𝑥→∞ 𝑥−2 𝑥2−5𝑥+6 = (𝑥−2) (𝑥−2)(𝑥−3) = 1 (𝑥−3) = 1 ∞ = 0 7. lim 𝑥→∞ 𝑥5+6𝑥−7 5𝑥6+6𝑥2−11 = 𝑥5+6𝑥3−7 𝑥6 5𝑥6+6𝑥2−11 𝑥6 = 1 𝑥 + 6 𝑥3− 7 𝑥6 5+ 6 𝑥4− 11 𝑥6 = 1 ∞ + 6 ∞ + 7 ∞ 5+ 6 ∞ − 11 ∞ = 0 5 = 0 8. lim 𝑥→∞ 7𝑥4+6𝑥2−3𝑥 −3𝑥3−7𝑥+5 = 7𝑥4+6𝑥2−3𝑥 𝑥4 −3𝑥3−7𝑥+5 𝑥4 = 7+ 6 𝑥2− 3 𝑥3 − 3 𝑥 − 7 𝑥3+ 5 𝑥4 = 7+ 6 ∞ − 3 ∞ − 3 ∞ − 7 ∞ + 5 ∞ = 7 0 = ∞ 9. lim 𝑥→∞ 2𝑥3+8𝑥−5 −3𝑥2+4 = 2𝑥3+8𝑥−5 𝑥3 −3𝑥2+4 𝑥3 = 2+ 8 𝑥2− 5 𝑥3 −3 𝑥 + 4 𝑥3 = 2+ 8 ∞ − 5 ∞ −3 ∞ + 4 ∞ = 2 0 = ∞ 10. lim 𝑥→∞ 5 𝑥2−4 = 5 𝑥2 𝑥2−4 𝑥2 = 5 𝑥2 1− 4 𝑥2 = 5 ∞ 1− 4 ∞ = 0 1 = 0
  3. 3. Latihan 2.3 1. lim 𝑥→4+ [ 𝑥] + 1 = 4 + 1 = 5 2. lim 𝑥→2− 𝑥2−4 𝑥−2 = ( 𝑥−2)(𝑥+2) 𝑥−2 = 2 + 2 = 4 3. lim 𝑥→8+ 4 𝑥−9 = 4 [8−9] = 4 −1 = 4 4. lim 𝑥→0+ √4𝑥 + 3 = √4(0) + 3 = √3 5. lim 𝑥→5 ( 𝑥 − 1) = 5 − 1 = 4 6. lim 𝑥→3− 𝑥2−9 𝑥−3 = ( 𝑥−3)( 𝑥+3) 𝑥−3 = 3 + 3 = −6 7. lim 𝑥→4+ 7 𝑥−4 = 7 4−4 = 7 0 = ∞ 8. lim 𝑥→4− 𝑥5+𝑥4−8 𝑥+4 = (−4)5+(−4)4−8 −4+4 = −1024+256−8 0 = −776 9. lim 𝑥→4+ 𝑥2−16 𝑥−4 = ( 𝑥−4)(𝑥+4) (𝑥−4) = 4 + 4 = 8 10. lim 𝑥→4− 𝑥2−16 𝑥−4 = ( 𝑥 + 4) = 4− + 4 = −8
  4. 4. Latihan 3.1 1. 𝑓( 𝑥) = √5𝑥 − 7 𝑑𝑖 𝑥 = 1 𝑓(1) = √−2 lim 𝑥→1 √5𝑥 − 7 = √−2 ↔ 𝑘𝑜𝑛𝑡𝑖𝑛𝑢 2. 𝑓( 𝑥) = 𝑥3−8 2−𝑥 𝑑𝑖 𝑥 = 0 𝑓(0) = −8 2 = −4 lim 𝑥→0 𝑥3−8 2−𝑥 = −4 ↔ 𝑘𝑜𝑛𝑡𝑖𝑛𝑢 3. 𝑓( 𝑥) = 4 √2𝑥−3 𝑑𝑖 𝑥 = 1 𝑓(1) = 4 √−1 lim 𝑥→0 4 √2𝑥−3 = 4 √−1 ↔ 𝑘𝑜𝑛𝑡𝑖𝑛𝑢 4. 𝑓( 𝑥) = [ 𝑥] 𝑑𝑖 𝑥 = 3 𝑓(3) = 3 lim 𝑥→0 [ 𝑥] = 3 ↔ 𝑘𝑜𝑛𝑡𝑖𝑛𝑢 5. 𝑔( 𝑥) = 𝑥2+6 𝑥−5 𝑑𝑖 𝑥 = 4 𝑓(4) = 10 −1 = −10 lim 𝑥→4 𝑥2+6 𝑥−5 = −10 ↔ 𝑘𝑜𝑛𝑡𝑖𝑛𝑢 6. 𝑔( 𝑥) = √𝑥−5 𝑥−5 𝑑𝑖 𝑥 = 3 𝑔(3) = √−2 −2 lim 𝑥→3 √𝑥−5 𝑥−5 = √−2 −2 ↔ 𝑘𝑜𝑛𝑡𝑖𝑛𝑢 7. 𝑔( 𝑥) = √𝑥−5 𝑥+2 𝑑𝑖 𝑥 = 8
  5. 5. 𝑔(8) = 2√2−5 10 lim 𝑥→8 √𝑥−5 𝑥+2 = 2√2−5 10 ↔ 𝑘𝑜𝑛𝑡𝑖𝑛𝑢 8. ℎ( 𝑥) = 5𝑥2 − √ 𝑥 + 7 𝑑𝑖 𝑥 = 5 ℎ(5) = 125 − √5 + 7 = 132 − √5 lim 𝑥→5 5𝑥2 − √ 𝑥 + 7 = 132 − √5 ↔ 𝑘𝑜𝑛𝑡𝑖𝑛𝑢 9. 𝑓( 𝑥) = 𝑥−6 𝑥−2 𝑑𝑖 𝑥 = 6 𝑓(6) = 0 4 = 0 lim 𝑥→6 𝑥−6 𝑥−2 = 0 ↔ 𝑘𝑜𝑛𝑡𝑖𝑛𝑢 10. ℎ( 𝑥) = (𝑥−𝑎)2+𝑥−6 𝑥−𝑎+3 𝑑𝑖 𝑥 = 𝑎 ℎ( 𝑎) = 0+𝑎−6 0+3 = 𝑎−6 3 lim 𝑥→𝑎 (𝑥−𝑎)2+𝑥−6 𝑥−𝑎+3 = 𝑎−6 3 ↔ 𝑘𝑜𝑛𝑡𝑖𝑛𝑢

×