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Teorema Limit ~ Matematika SMA | Catatan Nurkholis

Teorema Limit ~ Matematika SMA

Sebelumnya kita mengetahui bahwa, jika $f(a)$ terdefinisi, maka $displaystylelim_{xto a}f(x) = f(a)$. Dari sini, dapat dikembangkan menjadi beberapa teorema seperti berikut.

Teorema

Misalkan $n$ merupakan bilangan asli, $k$ konstanta, serta $f$ dan $g$ fungsi-fungsi yang mempunyai limit di $a$, maka

  1. $displaystylelim_{xto a}k = k$
    Contoh:$displaystylelim_{xto5}3 = 3$
  2. $displaystylelim_{xto a}kf(x) = klim_{xto a}f(x)$
    Contoh:
    $displaystylelim_{xto 2}5(x+7)= 5lim_{xto 2}(x+7)=5(9)=45$
  3. $displaystylelim_{xto a}left[f(x)pm g(x)right] = lim_{xto a}f(x)pmlim_{xto a}g(x)$
    Contoh:
    $displaystyle
    begin{aligned}
    lim_{xto 2}left[(x+7)+(2x-5)right]&=lim_{xto 2}(x+7)+lim_{xto 2}(2x-5)\
    lim_{xto 2} 3x+2 &=lim_{xto 2}(x+7)+lim_{xto 2}(2x-5)\
    3(2)+2 &= (2+7) + (2(2)-5)\
    8 &= 9 + (-1)\
    8 &= 8end{aligned}$
  4. $displaystylelim_{xto a}left[f(x)times g(x)right] = lim_{xto a}f(x)timeslim_{xto a}g(x)$
    Contoh:
    $displaystyle
    begin{aligned}
    lim_{xto 2}left[(x+7)(2x-5)right]&=lim_{xto 2}(x+7)timeslim_{xto 2}(2x-5)\
    lim_{xto 2} 2x^2+9x-35 &=lim_{xto 2}(x+7)timeslim_{xto 2}(2x-5)\
    2(2)^2+9(2)-35 &= (2+7) times (2(2)-5)\
    8+18-35 &= 9 times (-1)\
    -9 &= -9end{aligned}$
  5. $displaystylelim_{xto a}frac{f(x)}{g(x)} = frac{lim_{xto a}f(x)}{lim_{xto a}g(x)}$ dengan syarat $displaystylelim_{xto a}g(x)ne0$
    Contoh:
    $displaystyle
    begin{aligned}
    lim_{xto 2}frac{x+7}{2x-5}&=frac{lim_{xto 2}(x+7)}{lim_{xto 2}(2x-5)}\
    frac{2+7}{2(2)-5} &= frac{2+7}{2(2)-5} \
    -9 &= -9end{aligned}$
  6. $displaystylelim_{xto a}left[f(x)right]^n= left[lim_{xto a}f(x)right]^n$
    Contoh:
    $displaystyle
    begin{aligned}
    lim_{xto 2}(x+7)^2&=left(lim_{xto 2}(x+7)right)^2\
    lim_{xto 2}x^2+14x+49&=left(lim_{xto 2}(x+7)right)^2\
    (2)^2+14(2)+49 &= (2+7)^2 \
    81 &= 81end{aligned}$
  7. $displaystylelim_{xto a}sqrt[n]{f(x)}= sqrt[n]{lim_{xto a}f(x)}$
    Contoh:
    $displaystyle
    begin{aligned}
    lim_{xto 2}sqrt{x+7}&=sqrt{lim_{xto 2}(x+7)}\
    sqrt{9} &= sqrt{9} \
    3 &= 3end{aligned}$

Latihan Soal

Nah, sekarang latihan soal.

  1. Hitunglah!
    • $displaystylelim_{xto3}(x+3)(x+2)^2$
    • $displaystylelim_{xto3}sqrt{(x+3)^2-(x+2)^2}$
  2. Jika diketahui $displaystylelim_{xto a}f(x)=1$ dan $displaystylelim_{xto a}g(x)=3$. Tentukan nilai dari
    • $displaystyle lim_{xto a} left[f(x) + 2g(x)right]$
    • $displaystyle lim_{xto a} left[3f(x) – g(x)right]^5$
    • $displaystyle lim_{xto a} sqrt{f(x)^2 + g(x)^2}$

    Selamat Belajar!

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