Cara 2 Pembahasan Limit UM UGM 2009 Matematika IPA

Soal yang Akan Dibahas
$ \displaystyle \lim_{x \to \frac{\pi}{4} } \frac{\frac{1}{\sqrt{2}} \sin \left(\frac{\pi}{4} - 2x\right) + \frac{1}{\sqrt{2}} \cos \left(\frac{\pi}{4} - 2x\right) }{4x - \pi} = .... $
A). $ \frac{1}{4} \, $ B). $ \frac{1}{2} \, $ C). $ 0 \, $ D). $ -\frac{1}{4} \, $ E). $ -\frac{1}{2} $

$\spadesuit $ Konsep Dasar
*). Penerapan Turunan pada Limit (L'Hopital) :
$ \displaystyle \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0} \, $, solusinya $ \displaystyle \lim_{x \to a} \frac{f(x)}{g(x)} = \displaystyle \lim_{x \to a} \frac{f^\prime (x) }{g^\prime (x)} $
*). Turunan Fungsi trigonometri :
$ y = \sin f(x) \rightarrow y^\prime = f^\prime (x) \cos f(x) $
$ y = \cos f(x) \rightarrow y^\prime = -f^\prime (x) \sin f(x) $
*). Sifat sudut negatif : $ \sin (-x) = -\sin x \, $ dan $ \cos (-x) = \cos x $

$\clubsuit $ Pembahasan
*). Menentukan turunan fungsi trigonometrinya :
$ \begin{align} y & = \frac{1}{\sqrt{2}} \sin \left(\frac{\pi}{4} - 2x\right) \rightarrow y^\prime = \frac{-2}{\sqrt{2}} \cos \left(\frac{\pi}{4} - 2x\right) \\ y & = \frac{1}{\sqrt{2}} \cos \left(\frac{\pi}{4} - 2x\right) \rightarrow y^\prime = \frac{2}{\sqrt{2}} \sin \left(\frac{\pi}{4} - 2x\right) \end{align} $
*). Menyelesaikan soal :
$ \begin{align} & \displaystyle \lim_{x \to \frac{\pi}{4} } \frac{\frac{1}{\sqrt{2}} \sin \left(\frac{\pi}{4} - 2x\right) + \frac{1}{\sqrt{2}} \cos \left(\frac{\pi}{4} - 2x\right) }{4x - \pi} = \frac{0}{0} \\ & = \displaystyle \lim_{x \to \frac{\pi}{4} } \frac{\frac{-2}{\sqrt{2}} \cos \left(\frac{\pi}{4} - 2x\right) + \frac{2}{\sqrt{2}} \sin \left(\frac{\pi}{4} - 2x\right)}{4} \\ & = \frac{\frac{-2}{\sqrt{2}} \cos \left(\frac{\pi}{4} - 2. \frac{\pi}{4} \right) + \frac{2}{\sqrt{2}} \sin \left(\frac{\pi}{4} - 2. \frac{\pi}{4} \right)}{4} \\ & = \frac{\frac{-2}{\sqrt{2}} \cos \left(-\frac{\pi}{4} \right) + \frac{2}{\sqrt{2}} \sin \left(-\frac{\pi}{4} \right)}{4} \\ & = \frac{\frac{-2}{\sqrt{2}} \cos \left(\frac{\pi}{4} \right) - \frac{2}{\sqrt{2}} \sin \left(\frac{\pi}{4} \right)}{4} \\ & = \frac{\frac{-2}{\sqrt{2}} .\frac{1}{2}\sqrt{2} - \frac{2}{\sqrt{2}} . \frac{1}{2}\sqrt{2} }{4} \\ & = \frac{-1 -1 }{4} = \frac{-2}{4} = -\frac{1}{2} \end{align} $
Jadi, hasil limitnya adalah $ -\frac{1}{2} . \, \heartsuit $


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