Soal yang Akan Dibahas
$ \displaystyle \lim_{x \to \frac{\pi}{2} } \frac{\pi(\pi-2x)\tan \left(x - \frac{\pi}{2}\right)}{2(x-\pi)\cos ^2 x} = ..... $
A). $ -2 \, $ B). $ -1 \, $ C). $ -\frac{1}{2} \, $ D). $ 1 \, $ E). $ 2 $
A). $ -2 \, $ B). $ -1 \, $ C). $ -\frac{1}{2} \, $ D). $ 1 \, $ E). $ 2 $
$\spadesuit $ Konsep Dasar
*). Hubungan kuadrat :
$ \cos x = \sin \left( \frac{\pi}{2} - x \right) $
*). Sifat-sifat limit trigonometri :
$ \displaystyle \lim_{x \to k } \frac{\tan af(x)}{bf(x)} = \frac{a}{b} $ dan $ \displaystyle \lim_{x \to k } \frac{\tan af(x)}{\sin bf(x)} = \frac{a}{b} $ dengan syarat $ f(k) = 0 $
*). Hubungan kuadrat :
$ \cos x = \sin \left( \frac{\pi}{2} - x \right) $
*). Sifat-sifat limit trigonometri :
$ \displaystyle \lim_{x \to k } \frac{\tan af(x)}{bf(x)} = \frac{a}{b} $ dan $ \displaystyle \lim_{x \to k } \frac{\tan af(x)}{\sin bf(x)} = \frac{a}{b} $ dengan syarat $ f(k) = 0 $
$\clubsuit $ Pembahasan
*). Menyelesaikan soal :
$\begin{align} & \displaystyle \lim_{x \to \frac{\pi}{2} } \frac{\pi(\pi-2x)\tan \left(x - \frac{\pi}{2}\right)}{2(x-\pi)\cos ^2 x} \\ & = \displaystyle \lim_{x \to \frac{\pi}{2} } \frac{\pi(\pi-2x)\tan \left(x - \frac{\pi}{2}\right)}{2(x-\pi)\sin ^2 \left( \frac{\pi}{2} - x \right) } \\ & = \displaystyle \lim_{x \to \frac{\pi}{2} } \frac{\pi.2\left( \frac{\pi}{2} - x \right)\tan -\left( \frac{\pi}{2} - x \right)}{2(x-\pi)\sin \left( \frac{\pi}{2} - x \right) . \sin \left( \frac{\pi}{2} - x \right) } \\ & = \displaystyle \lim_{x \to \frac{\pi}{2} } \frac{2\pi}{2(x-\pi)} . \frac{\left( \frac{\pi}{2} - x \right)}{\sin \left( \frac{\pi}{2} - x \right) } . \frac{\tan -\left( \frac{\pi}{2} - x \right)}{ \sin \left( \frac{\pi}{2} - x \right) } \\ & = \frac{\pi}{\frac{\pi}{2} -\pi} . \frac{1}{1} . \frac{-1}{ 1 } \\ & = \frac{\pi}{-\frac{\pi}{2} } . 1 . -1 \\ & = -2 . 1 . -1 = 2 \end{align} $
Jadi, nilai limitnya adalah $ 2 . \, \heartsuit $
*). Menyelesaikan soal :
$\begin{align} & \displaystyle \lim_{x \to \frac{\pi}{2} } \frac{\pi(\pi-2x)\tan \left(x - \frac{\pi}{2}\right)}{2(x-\pi)\cos ^2 x} \\ & = \displaystyle \lim_{x \to \frac{\pi}{2} } \frac{\pi(\pi-2x)\tan \left(x - \frac{\pi}{2}\right)}{2(x-\pi)\sin ^2 \left( \frac{\pi}{2} - x \right) } \\ & = \displaystyle \lim_{x \to \frac{\pi}{2} } \frac{\pi.2\left( \frac{\pi}{2} - x \right)\tan -\left( \frac{\pi}{2} - x \right)}{2(x-\pi)\sin \left( \frac{\pi}{2} - x \right) . \sin \left( \frac{\pi}{2} - x \right) } \\ & = \displaystyle \lim_{x \to \frac{\pi}{2} } \frac{2\pi}{2(x-\pi)} . \frac{\left( \frac{\pi}{2} - x \right)}{\sin \left( \frac{\pi}{2} - x \right) } . \frac{\tan -\left( \frac{\pi}{2} - x \right)}{ \sin \left( \frac{\pi}{2} - x \right) } \\ & = \frac{\pi}{\frac{\pi}{2} -\pi} . \frac{1}{1} . \frac{-1}{ 1 } \\ & = \frac{\pi}{-\frac{\pi}{2} } . 1 . -1 \\ & = -2 . 1 . -1 = 2 \end{align} $
Jadi, nilai limitnya adalah $ 2 . \, \heartsuit $