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# Tugas Calculus : Limit (Hal. 8-14)

Tugas Calculus : Limit (Hal. 8-14)

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### 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 π₯ββ (5π₯ β 7) = β 2. lim π₯ββ 7 π₯3 = 7 β = 0 3. lim π₯ββ 3π₯ + 95 = β + 95 = β 4. lim π₯ββ π₯3βπ₯2+47π₯+9 18π₯3+76π₯β11 = 1 18 5. lim π₯ββ 8 4βπ₯ = 8 β = 0 6. lim π₯ββ π₯β2 π₯2β5π₯+6 = (π₯β2) (π₯β2)(π₯β3) = 1 (π₯β3) = 1 β = 0 7. lim π₯ββ π₯5+6π₯β7 5π₯6+6π₯2β11 = π₯5+6π₯3β7 π₯6 5π₯6+6π₯2β11 π₯6 = 1 π₯ + 6 π₯3β 7 π₯6 5+ 6 π₯4β 11 π₯6 = 1 β + 6 β + 7 β 5+ 6 β β 11 β = 0 5 = 0 8. lim π₯ββ 7π₯4+6π₯2β3π₯ β3π₯3β7π₯+5 = 7π₯4+6π₯2β3π₯ π₯4 β3π₯3β7π₯+5 π₯4 = 7+ 6 π₯2β 3 π₯3 β 3 π₯ β 7 π₯3+ 5 π₯4 = 7+ 6 β β 3 β β 3 β β 7 β + 5 β = 7 0 = β 9. lim π₯ββ 2π₯3+8π₯β5 β3π₯2+4 = 2π₯3+8π₯β5 π₯3 β3π₯2+4 π₯3 = 2+ 8 π₯2β 5 π₯3 β3 π₯ + 4 π₯3 = 2+ 8 β β 5 β β3 β + 4 β = 2 0 = β 10. lim π₯ββ 5 π₯2β4 = 5 π₯2 π₯2β4 π₯2 = 5 π₯2 1β 4 π₯2 = 5 β 1β 4 β = 0 1 = 0
3. 3. Latihan 2.3 1. lim π₯β4+ [ π₯] + 1 = 4 + 1 = 5 2. lim π₯β2β π₯2β4 π₯β2 = ( π₯β2)(π₯+2) π₯β2 = 2 + 2 = 4 3. lim π₯β8+ 4 π₯β9 = 4 [8β9] = 4 β1 = 4 4. lim π₯β0+ β4π₯ + 3 = β4(0) + 3 = β3 5. lim π₯β5 ( π₯ β 1) = 5 β 1 = 4 6. lim π₯β3β π₯2β9 π₯β3 = ( π₯β3)( π₯+3) π₯β3 = 3 + 3 = β6 7. lim π₯β4+ 7 π₯β4 = 7 4β4 = 7 0 = β 8. lim π₯β4β π₯5+π₯4β8 π₯+4 = (β4)5+(β4)4β8 β4+4 = β1024+256β8 0 = β776 9. lim π₯β4+ π₯2β16 π₯β4 = ( π₯β4)(π₯+4) (π₯β4) = 4 + 4 = 8 10. lim π₯β4β π₯2β16 π₯β4 = ( π₯ + 4) = 4β + 4 = β8
4. 4. Latihan 3.1 1. π( π₯) = β5π₯ β 7 ππ π₯ = 1 π(1) = ββ2 lim π₯β1 β5π₯ β 7 = ββ2 β ππππ‘πππ’ 2. π( π₯) = π₯3β8 2βπ₯ ππ π₯ = 0 π(0) = β8 2 = β4 lim π₯β0 π₯3β8 2βπ₯ = β4 β ππππ‘πππ’ 3. π( π₯) = 4 β2π₯β3 ππ π₯ = 1 π(1) = 4 ββ1 lim π₯β0 4 β2π₯β3 = 4 ββ1 β ππππ‘πππ’ 4. π( π₯) = [ π₯] ππ π₯ = 3 π(3) = 3 lim π₯β0 [ π₯] = 3 β ππππ‘πππ’ 5. π( π₯) = π₯2+6 π₯β5 ππ π₯ = 4 π(4) = 10 β1 = β10 lim π₯β4 π₯2+6 π₯β5 = β10 β ππππ‘πππ’ 6. π( π₯) = βπ₯β5 π₯β5 ππ π₯ = 3 π(3) = ββ2 β2 lim π₯β3 βπ₯β5 π₯β5 = ββ2 β2 β ππππ‘πππ’ 7. π( π₯) = βπ₯β5 π₯+2 ππ π₯ = 8
5. 5. π(8) = 2β2β5 10 lim π₯β8 βπ₯β5 π₯+2 = 2β2β5 10 β ππππ‘πππ’ 8. β( π₯) = 5π₯2 β β π₯ + 7 ππ π₯ = 5 β(5) = 125 β β5 + 7 = 132 β β5 lim π₯β5 5π₯2 β β π₯ + 7 = 132 β β5 β ππππ‘πππ’ 9. π( π₯) = π₯β6 π₯β2 ππ π₯ = 6 π(6) = 0 4 = 0 lim π₯β6 π₯β6 π₯β2 = 0 β ππππ‘πππ’ 10. β( π₯) = (π₯βπ)2+π₯β6 π₯βπ+3 ππ π₯ = π β( π) = 0+πβ6 0+3 = πβ6 3 lim π₯βπ (π₯βπ)2+π₯β6 π₯βπ+3 = πβ6 3 β ππππ‘πππ’