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Pembuktian Aturan L’hopital Kasus Tak Hingga / Tak Hingga

  • Tulis  x=\frac{1}{1/x}, f(x)=\frac{1}{1/f(x)} dan g(x)=\frac{1}{1/g(x)}.
  • Misal

    \[ \latex \lim_{x \rightarrow a} \frac{f(x)}{g(x)}= \frac {\infty}{\infty}\]

dan limit di atas ada, yaitu L

  • Maka

    \[ \latex \lim_{x \rightarrow a} \frac{f(x)}{g(x)}= \lim_{x \rightarrow a} \frac{\frac{1}{1/f(x)}}{\frac{1}{1/g(x)}}=\lim_{x \rightarrow a} \frac{     \frac{1}{g(x)}        }{        \frac{1}{f(x)}     }=\frac{0}{0}\]

Nah….jadi bentuk 0/0 kan? hayo kalau masih ngaa… inget \lim_{x \rightarrow \infty} \frac{1}{x}=0


    \[ \latex \lim_{x \rightarrow a} \frac{f(x)}{g(x)}= \lim_{x \rightarrow a} \frac{\frac{1}{1/f(x)}}{\frac{1}{1/g(x)}}=\lim_{x \rightarrow a} \frac{     \frac{1}{g(x)}        }{        \frac{1}{f(x)}     }=\frac{0}{0}=\lim_{x \rightarrow a} \frac{(\frac{1}{g(x)})'}{(\frac{1}{f(x)})'}\]

    \[ \latex \lim_{x \rightarrow a} \frac{f(x)}{g(x)}= \frac{\frac{-g'(x)}{g^2(x)}}{\frac{-f'(x)}{f^2(x)}}\]

    \[ \latex \lim_{x \rightarrow a} \frac{f(x)}{g(x)}= \frac{f^2(x).g'(x)}{g^2(x).f'(x) }\]

    \[ \latex \lim_{x \rightarrow a} \frac{f(x)}{g(x)}= \lim_{x \rightarrow a} \frac{f^2(x)}{g^2(x) }\lim_{x \rightarrow a} \frac{g'(x)}{f'(x)}\]

    \[ \latex \lim_{x \rightarrow a} \frac{f(x)}{g(x)}= \frac{L^2}{\lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}}\]

Seandainya jika limitnya ada atau dengan kata lain

    \[\latex \lim_{x \rightarrow a} \frac{f(x)}{g(x)}=L\]


    \[L=\lim_{x \rightarrow a} \frac{f(x)}{g(x)}= \frac{L^2}{\lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}}\]

atau jika bagian tengah “=” nya kita hilangin,

    \[L= \frac{L^2}{\lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}}\]

sehingga bagian penyebut haruslah L, yang mana


    \[ \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}} = L \]

kesimpulannya adalah

    \[ \latex \lim_{x \rightarrow a} \frac{f(x)}{g(x)}=\lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}} = L \]


Catatan : pembuktian ini untuk L bukan nol dan bukan takhingga.

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