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
*). Penyelesaian limit dengan turunan (L'Hopital) :
$ \displaystyle \lim_{x \to k} \frac{ f(x)}{g(x)} = \frac{0}{0} \, $ memiliki solusi $ \displaystyle \lim_{x \to k} \frac{ f(x)}{g(x)} = \displaystyle \lim_{x \to k} \frac{ f^\prime (x)}{g^\prime (x)} $
*). 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 $
*). Turunan $ y = \tan ax \rightarrow y^\prime = a\sec ^2 ax $
*). Penyelesaian limit dengan turunan (L'Hopital) :
$ \displaystyle \lim_{x \to k} \frac{ f(x)}{g(x)} = \frac{0}{0} \, $ memiliki solusi $ \displaystyle \lim_{x \to k} \frac{ f(x)}{g(x)} = \displaystyle \lim_{x \to k} \frac{ f^\prime (x)}{g^\prime (x)} $
*). 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 $
*). Turunan $ y = \tan ax \rightarrow y^\prime = a\sec ^2 ax $
$\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 } \, \, \, \, \, \, \text{(L'Hopital)} \\ & = \left( 1- \sin \left(0 - \frac{\pi}{2} \right) \right) . \displaystyle \lim_{x \to 0} \, \frac{1}{2\sec ^2 2x } \\ & = \left( 1- \sin \left( - \frac{\pi}{2} \right) \right) . \frac{1}{2\sec ^2 0 } \\ & = \left( 1+ \sin \left( \frac{\pi}{2} \right) \right) . \frac{1}{2 . 1} \\ & = \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 } \, \, \, \, \, \, \text{(L'Hopital)} \\ & = \left( 1- \sin \left(0 - \frac{\pi}{2} \right) \right) . \displaystyle \lim_{x \to 0} \, \frac{1}{2\sec ^2 2x } \\ & = \left( 1- \sin \left( - \frac{\pi}{2} \right) \right) . \frac{1}{2\sec ^2 0 } \\ & = \left( 1+ \sin \left( \frac{\pi}{2} \right) \right) . \frac{1}{2 . 1} \\ & = \left( 1+ 1 \right) . \frac{1}{2} = 2 . \frac{1}{2} = 1 \end{align} $
Jadi, hasil limitnya adalah $ 1 . \, \heartsuit $
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