Rabu, 19 Februari 2014

Tugas XI IPS 2



Hitunglah tiap limit berikut dengan menggunakan teorema limit
11.       Lim (2x3 + 3x – 6)
   X -> -1
22.       Lim (3x2 + x) / (akar x4 + 3)
   X -> 1
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9 komentar:

  1. Unknown22 Februari 2014 pukul 00.25

    Komentar ini telah dihapus oleh pengarang.

    BalasHapus
  2. Unknown22 Februari 2014 pukul 00.29

    Jawaban:
    lim/(x→1) (2x2+3x-6)=teorema 4
    =lim/(x→1)(2(x)2+lim/(x→1)3(x)-6
    =teorema 3
    =2lim/(x→1) x2+lim/(x→3) 3X-6
    Teorema 8
    =2( lim/(x→1 ) x)2+ lim/(x→1) 3x-6
    Teorema 2
    =2(1)2+lim/(x→1) 3x-6
    Teorema 5
    =2(1)2 +(lim/(x→1) 3(x) - lim/(x→1) 6)
    Teorema 3
    =2(1)2 + (3lim/(x→1) x - lim/(x→1) 6)
    Teorema 2
    =2(1)2+(3.1-6)
    Teorema 1
    2 + 3 – 6 =2 + (-3) = -1

    lim/(x→1) ((3x^2+x))/((akar x^4+3)) 7=lim 3(x)2+x untuk x mndekati
    2= (3(1)^2+1)/(akar lim⁡〖(x)^4 〗+3 untuk x mendekati) =(3+1)/(akar lim⁡〖(x)^4 〗+3)=4/(akar lim⁡〖(x)^4 〗+3) lim4/(akar lim⁡〖(x)^4 〗+3untk x mndkati)=9=4/(akar lim⁡〖(x)^4 〗+3)=4/(akar lim⁡〖x^4 〗+lim⁡〖3 〗 untk x mndkti)
    24/(akar (1)^4+3)=4/(akar 4)=4/2=2


    BalasHapus
  3. Unknown22 Februari 2014 pukul 00.31

    pak gru yg d'bwa skli nmr 2...

    BalasHapus
  4. Unknown22 Februari 2014 pukul 00.34

    Jawaban:
    lim/(x→1) (2x2+3x-6)=teorema 4
    =lim/(x→1)(2(x)2+lim/(x→1)3(x)-6
    =teorema 3
    =2lim/(x→1) x2+lim/(x→3) 3X-6
    Teorema 8
    =2( lim/(x→1 ) x)2+ lim/(x→1) 3x-6
    Teorema 2
    =2(1)2+lim/(x→1) 3x-6
    Teorema 5
    =2(1)2 +(lim/(x→1) 3(x) - lim/(x→1) 6)
    Teorema 3
    =2(1)2 + (3lim/(x→1) x - lim/(x→1) 6)
    Teorema 2
    =2(1)2+(3.1-6)
    Teorema 1
    2 + 3 – 6 =2 + (-3) = -1

    lim/(x→1) ((3x^2+x))/((akar x^4+3)) 7=lim 3(x)2+x untuk x mndekati
    2= (3(1)^2+1)/(akar lim⁡〖(x)^4 〗+3 untuk x mendekati) =(3+1)/(akar lim⁡〖(x)^4 〗+3)=4/(akar lim⁡〖(x)^4 〗+3) lim4/(akar lim⁡〖(x)^4 〗+3untk x mndkati)=9=4/(akar lim⁡〖(x)^4 〗+3)=4/(akar lim⁡〖x^4 〗+lim⁡〖3 〗 untk x mndkti)
    24/(akar (1)^4+3)=4/(akar 4)=4/2=2


    BalasHapus
  5. Unknown22 Februari 2014 pukul 00.37

    Jawaban:
    lim/(x→1) (2x2+3x-6)=teorema 4
    =lim/(x→1)(2(x)2+lim/(x→1)3(x)-6
    =teorema 3
    =2lim/(x→1) x2+lim/(x→3) 3X-6
    Teorema 8
    =2( lim/(x→1 ) x)2+ lim/(x→1) 3x-6
    Teorema 2
    =2(1)2+lim/(x→1) 3x-6
    Teorema 5
    =2(1)2 +(lim/(x→1) 3(x) - lim/(x→1) 6)
    Teorema 3
    =2(1)2 + (3lim/(x→1) x - lim/(x→1) 6)
    Teorema 2
    =2(1)2+(3.1-6)
    Teorema 1
    2 + 3 – 6 =2 + (-3) = -1

    lim/(x→1) ((3x^2+x))/((akar x^4+3)) 7=lim 3(x)2+x untuk x mndekati
    2= (3(1)^2+1)/(akar lim⁡〖(x)^4 〗+3 untuk x mendekati) =(3+1)/(akar lim⁡〖(x)^4 〗+3)=4/(akar lim⁡〖(x)^4 〗+3) lim4/(akar lim⁡〖(x)^4 〗+3untk x mndkati)=9=4/(akar lim⁡〖(x)^4 〗+3)=4/(akar lim⁡〖x^4 〗+lim⁡〖3 〗 untk x mndkti)
    24/(akar (1)^4+3)=4/(akar 4)=4/2=2


    BalasHapus
  6. Unknown22 Februari 2014 pukul 00.40

    Jawaban:
    lim/(x→1) (2x2+3x-6)=teorema 4
    =lim/(x→1)(2(x)2+lim/(x→1)3(x)-6
    =teorema 3
    =2lim/(x→1) x2+lim/(x→3) 3X-6
    Teorema 8
    =2( lim/(x→1 ) x)2+ lim/(x→1) 3x-6
    Teorema 2
    =2(1)2+lim/(x→1) 3x-6
    Teorema 5
    =2(1)2 +(lim/(x→1) 3(x) - lim/(x→1) 6)
    Teorema 3
    =2(1)2 + (3lim/(x→1) x - lim/(x→1) 6)
    Teorema 2
    =2(1)2+(3.1-6)
    Teorema 1
    2 + 3 – 6 =2 + (-3) = -1

    lim/(x→1) ((3x^2+x))/((akar x^4+3)) 7=lim 3(x)2+x untuk x mndekati
    2= (3(1)^2+1)/(akar lim⁡〖(x)^4 〗+3 untuk x mendekati) =(3+1)/(akar lim⁡〖(x)^4 〗+3)=4/(akar lim⁡〖(x)^4 〗+3) lim4/(akar lim⁡〖(x)^4 〗+3untk x mndkati)=9=4/(akar lim⁡〖(x)^4 〗+3)=4/(akar lim⁡〖x^4 〗+lim⁡〖3 〗 untk x mndkti)
    24/(akar (1)^4+3)=4/(akar 4)=4/2=2


    BalasHapus
  7. Unknown22 Februari 2014 pukul 00.42

    Jawaban:
    lim/(x→1) (2x2+3x-6)=teorema 4
    =lim/(x→1)(2(x)2+lim/(x→1)3(x)-6
    =teorema 3
    =2lim/(x→1) x2+lim/(x→3) 3X-6
    Teorema 8
    =2( lim/(x→1 ) x)2+ lim/(x→1) 3x-6
    Teorema 2
    =2(1)2+lim/(x→1) 3x-6
    Teorema 5
    =2(1)2 +(lim/(x→1) 3(x) - lim/(x→1) 6)
    Teorema 3
    =2(1)2 + (3lim/(x→1) x - lim/(x→1) 6)
    Teorema 2
    =2(1)2+(3.1-6)
    Teorema 1
    2 + 3 – 6 =2 + (-3) = -1

    lim/(x→1) ((3x^2+x))/((akar x^4+3)) 7=lim 3(x)2+x untuk x mndekati
    2= (3(1)^2+1)/(akar lim⁡〖(x)^4 〗+3 untuk x mendekati) =(3+1)/(akar lim⁡〖(x)^4 〗+3)=4/(akar lim⁡〖(x)^4 〗+3) lim4/(akar lim⁡〖(x)^4 〗+3untk x mndkati)=9=4/(akar lim⁡〖(x)^4 〗+3)=4/(akar lim⁡〖x^4 〗+lim⁡〖3 〗 untk x mndkti)
    24/(akar (1)^4+3)=4/(akar 4)=4/2=2


    BalasHapus
  8. Unknown22 Februari 2014 pukul 00.45

    Jawaban:
    lim/(x→1) (2x2+3x-6)=teorema 4
    =lim/(x→1)(2(x)2+lim/(x→1)3(x)-6
    =teorema 3
    =2lim/(x→1) x2+lim/(x→3) 3X-6
    Teorema 8
    =2( lim/(x→1 ) x)2+ lim/(x→1) 3x-6
    Teorema 2
    =2(1)2+lim/(x→1) 3x-6
    Teorema 5
    =2(1)2 +(lim/(x→1) 3(x) - lim/(x→1) 6)
    Teorema 3
    =2(1)2 + (3lim/(x→1) x - lim/(x→1) 6)
    Teorema 2
    =2(1)2+(3.1-6)
    Teorema 1
    2 + 3 – 6 =2 + (-3) = -1

    lim/(x→1) ((3x^2+x))/((akar x^4+3)) 7=lim 3(x)2+x untuk x mndekati
    2= (3(1)^2+1)/(akar lim⁡〖(x)^4 〗+3 untuk x mendekati) =(3+1)/(akar lim⁡〖(x)^4 〗+3)=4/(akar lim⁡〖(x)^4 〗+3) lim4/(akar lim⁡〖(x)^4 〗+3untk x mndkati)=9=4/(akar lim⁡〖(x)^4 〗+3)=4/(akar lim⁡〖x^4 〗+lim⁡〖3 〗 untk x mndkti)
    24/(akar (1)^4+3)=4/(akar 4)=4/2=2


    BalasHapus
  9. Anonim22 Februari 2014 pukul 01.10

    Jawaban:
    lim/(x→1) (2x2+3x-6)=teorema 4
    =lim/(x→1)(2(x)2+lim/(x→1)3(x)-6
    =teorema 3
    =2lim/(x→1) x2+lim/(x→3) 3X-6
    Teorema 8
    =2( lim/(x→1 ) x)2+ lim/(x→1) 3x-6
    Teorema 2
    =2(1)2+lim/(x→1) 3x-6
    Teorema 5
    =2(1)2 +(lim/(x→1) 3(x) - lim/(x→1) 6)
    Teorema 3
    =2(1)2 + (3lim/(x→1) x - lim/(x→1) 6)
    Teorema 2
    =2(1)2+(3.1-6)
    Teorema 1
    2 + 3 – 6 =2 + (-3) = -1

    lim/(x→1) ((3x^2+x))/((akar x^4+3)) 7=lim 3(x)2+x untuk x mndekati
    2= (3(1)^2+1)/(akar lim⁡〖(x)^4 〗+3 untuk x mendekati) =(3+1)/(akar lim⁡〖(x)^4 〗+3)=4/(akar lim⁡〖(x)^4 〗+3) lim4/(akar lim⁡〖(x)^4 〗+3untk x mndkati)=9=4/(akar lim⁡〖(x)^4 〗+3)=4/(akar lim⁡〖x^4 〗+lim⁡〖3 〗 untk x mndkti)
    24/(akar (1)^4+3)=4/(akar 4)=4/2=2



    BalasHapus

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