Soal yang Akan Dibahas
$ \displaystyle \lim_{x \to 0} \,
x.\left( 1- \sin \left(x - \frac{\pi}{2} \right) \right) . \cot ( 2x - \pi) = .... $
A). $ \frac{1}{2} \, $ B). $ 1 \, $ C). $ \frac{3}{2} \, $ D). $ 2 \, $ E). $ \frac{5}{2} $
A). $ \frac{1}{2} \, $ B). $ 1 \, $ C). $ \frac{3}{2} \, $ D). $ 2 \, $ E). $ \frac{5}{2} $
$\spadesuit $ Konsep Dasar
*). Sifat limit trigonometri :
$ \displaystyle \lim_{x \to 0} \frac{ ax}{\tan bx} = \frac{a}{b} \, $
*). Rumus dasar trigonometri :
$ \sin (- x) = - \sin x \, $ dan $ \tan ( - x) = -\tan x $
$ \tan (\pi - A) = -\tan A \, $ dan $ \cot A = \frac{1}{\tan A} $
$ \tan ( A - \pi) = \tan -(\pi - A) = -\tan (\pi - A) = \tan A $
*). Sifat limit trigonometri :
$ \displaystyle \lim_{x \to 0} \frac{ ax}{\tan bx} = \frac{a}{b} \, $
*). Rumus dasar trigonometri :
$ \sin (- x) = - \sin x \, $ dan $ \tan ( - x) = -\tan x $
$ \tan (\pi - A) = -\tan A \, $ dan $ \cot A = \frac{1}{\tan A} $
$ \tan ( A - \pi) = \tan -(\pi - A) = -\tan (\pi - A) = \tan A $
$\clubsuit $ Pembahasan
*). Menyelesaikan soal :
$\begin{align} & \displaystyle \lim_{x \to 0} \, x.\left( 1- \sin \left(x - \frac{\pi}{2} \right) \right) . \cot ( 2x - \pi) \\ & = \displaystyle \lim_{x \to 0} \, x.\left( 1- \sin \left(x - \frac{\pi}{2} \right) \right) . \frac{1}{\tan ( 2x - \pi) } \\ & = \displaystyle \lim_{x \to 0} \, \left( 1- \sin \left(x - \frac{\pi}{2} \right) \right) . \frac{x}{\tan ( 2x - \pi) } \\ & = \displaystyle \lim_{x \to 0} \, \left( 1- \sin \left(x - \frac{\pi}{2} \right) \right) . \frac{x}{\tan 2x } \\ & = \displaystyle \lim_{x \to 0} \, \left( 1- \sin \left(x - \frac{\pi}{2} \right) \right) . \displaystyle \lim_{x \to 0} \, \frac{x}{\tan 2x } \\ & = \left( 1- \sin \left(0 - \frac{\pi}{2} \right) \right) . \frac{1}{2} \\ & = \left( 1- \sin \left( - \frac{\pi}{2} \right) \right) . \frac{1}{2} \\ & = \left( 1+ \sin \left( \frac{\pi}{2} \right) \right) . \frac{1}{2} \\ & = \left( 1+ 1 \right) . \frac{1}{2} = 2 . \frac{1}{2} = 1 \end{align} $
Jadi, hasil limitnya adalah $ 1 . \, \heartsuit $
*). Menyelesaikan soal :
$\begin{align} & \displaystyle \lim_{x \to 0} \, x.\left( 1- \sin \left(x - \frac{\pi}{2} \right) \right) . \cot ( 2x - \pi) \\ & = \displaystyle \lim_{x \to 0} \, x.\left( 1- \sin \left(x - \frac{\pi}{2} \right) \right) . \frac{1}{\tan ( 2x - \pi) } \\ & = \displaystyle \lim_{x \to 0} \, \left( 1- \sin \left(x - \frac{\pi}{2} \right) \right) . \frac{x}{\tan ( 2x - \pi) } \\ & = \displaystyle \lim_{x \to 0} \, \left( 1- \sin \left(x - \frac{\pi}{2} \right) \right) . \frac{x}{\tan 2x } \\ & = \displaystyle \lim_{x \to 0} \, \left( 1- \sin \left(x - \frac{\pi}{2} \right) \right) . \displaystyle \lim_{x \to 0} \, \frac{x}{\tan 2x } \\ & = \left( 1- \sin \left(0 - \frac{\pi}{2} \right) \right) . \frac{1}{2} \\ & = \left( 1- \sin \left( - \frac{\pi}{2} \right) \right) . \frac{1}{2} \\ & = \left( 1+ \sin \left( \frac{\pi}{2} \right) \right) . \frac{1}{2} \\ & = \left( 1+ 1 \right) . \frac{1}{2} = 2 . \frac{1}{2} = 1 \end{align} $
Jadi, hasil limitnya adalah $ 1 . \, \heartsuit $
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