Soal yang Akan Dibahas
$ \displaystyle \lim_{x \to 0 }
\frac{x\tan 5x}{\cos 2x - \cos 7x } = .... $
A). $ \frac{1}{9} \, $ B). $ -\frac{1}{9} \, $ C). $ \frac{2}{9} \, $ D). $ -\frac{2}{9} \, $ E). $ 0 $
A). $ \frac{1}{9} \, $ B). $ -\frac{1}{9} \, $ C). $ \frac{2}{9} \, $ D). $ -\frac{2}{9} \, $ E). $ 0 $
$\spadesuit $ Konsep Dasar
*). Sifat limit trigonometri :
$ \displaystyle \lim_{x \to 0 } \frac{ax}{\sin bx} = \frac{a}{b} \, $ dan $ \displaystyle \lim_{x \to 0 } \frac{\tan ax}{\sin bx} = \frac{a}{b} $.
*). Rumus trigonometri :
i). $ \cos A - \cos B = -2\sin \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right) $
ii). $ \sin (-x ) = - \sin x $
*). Sifat limit trigonometri :
$ \displaystyle \lim_{x \to 0 } \frac{ax}{\sin bx} = \frac{a}{b} \, $ dan $ \displaystyle \lim_{x \to 0 } \frac{\tan ax}{\sin bx} = \frac{a}{b} $.
*). Rumus trigonometri :
i). $ \cos A - \cos B = -2\sin \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right) $
ii). $ \sin (-x ) = - \sin x $
$\clubsuit $ Pembahasan
*). Mengubah bentuk trigonometrinya :
$ \begin{align} \cos 2x - \cos 7x & = -2\sin \left( \frac{2x+7x}{2} \right) \sin \left( \frac{2x - 7x}{2} \right) \\ & = -2\sin \left( \frac{9}{2} x \right) \sin \left( -\frac{5}{2} x \right) \\ & = 2\sin \left( \frac{9}{2} x \right) \sin \left( \frac{5}{2} x \right) \end{align} $
*). Menyelesaikan limitnya :
$ \begin{align} & \displaystyle \lim_{x \to 0 } \frac{x\tan 5x}{\cos 2x - \cos 7x } \\ & = \displaystyle \lim_{x \to 0 } \frac{x\tan 5x}{ 2\sin \left( \frac{9}{2} x \right) \sin \left( \frac{5}{2} x \right) } \\ & = \displaystyle \lim_{x \to 0 } \frac{1}{2}. \frac{x}{\sin \left( \frac{9}{2} x \right)}. \frac{\tan 5x}{ \sin \left( \frac{5}{2} x \right) } \\ & = \frac{1}{2}. \frac{1}{ \frac{9}{2} }. \frac{ 5}{\frac{5}{2} } \\ & = \frac{1}{2}. \frac{2}{ 9}. 5 . \frac{ 2}{5} = \frac{2}{9} \end{align} $
Jadi, hasil limitnya adalah $ \frac{2}{9} . \, \heartsuit $
*). Mengubah bentuk trigonometrinya :
$ \begin{align} \cos 2x - \cos 7x & = -2\sin \left( \frac{2x+7x}{2} \right) \sin \left( \frac{2x - 7x}{2} \right) \\ & = -2\sin \left( \frac{9}{2} x \right) \sin \left( -\frac{5}{2} x \right) \\ & = 2\sin \left( \frac{9}{2} x \right) \sin \left( \frac{5}{2} x \right) \end{align} $
*). Menyelesaikan limitnya :
$ \begin{align} & \displaystyle \lim_{x \to 0 } \frac{x\tan 5x}{\cos 2x - \cos 7x } \\ & = \displaystyle \lim_{x \to 0 } \frac{x\tan 5x}{ 2\sin \left( \frac{9}{2} x \right) \sin \left( \frac{5}{2} x \right) } \\ & = \displaystyle \lim_{x \to 0 } \frac{1}{2}. \frac{x}{\sin \left( \frac{9}{2} x \right)}. \frac{\tan 5x}{ \sin \left( \frac{5}{2} x \right) } \\ & = \frac{1}{2}. \frac{1}{ \frac{9}{2} }. \frac{ 5}{\frac{5}{2} } \\ & = \frac{1}{2}. \frac{2}{ 9}. 5 . \frac{ 2}{5} = \frac{2}{9} \end{align} $
Jadi, hasil limitnya adalah $ \frac{2}{9} . \, \heartsuit $
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