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
*). Sifat limit fungsi trigonometri :
$ \displaystyle \lim_{x \to a} \frac{\sin a f(x)}{b f(x)} = \frac{a}{b} \, $ dengan syarat $ f(a) = 0 $
*). Rumus jumlah trigonometri :
$ \sin (A + B) = \sin A \cos B + \cos A \sin B $
*). Nilai $ \cos 45^\circ = \cos \frac{\pi}{4} = \frac{1}{2}\sqrt{2} = \frac{1}{\sqrt{2}} $
*). Nilai $ \sin 45^\circ = \sin \frac{\pi}{4} = \frac{1}{2}\sqrt{2} = \frac{1}{\sqrt{2}} $

$\clubsuit $ Pembahasan
*). Menyederhanakan soal :
$ \begin{align} & \frac{1}{\sqrt{2}} \sin \left(\frac{\pi}{4} - 2x\right) + \frac{1}{\sqrt{2}} \cos \left(\frac{\pi}{4} - 2x\right) \\ & = \sin \left(\frac{\pi}{4} - 2x\right).\frac{1}{\sqrt{2}} + \cos \left(\frac{\pi}{4} - 2x\right) .\frac{1}{\sqrt{2}} \\ & = \sin \left(\frac{\pi}{4} - 2x\right).\cos \frac{\pi}{4} + \cos \left(\frac{\pi}{4} - 2x\right) .\sin \frac{\pi}{4} \\ & = \sin \left( \left(\frac{\pi}{4} - 2x\right) + \frac{\pi}{4} \right) \\ & = \sin \left( \frac{\pi}{2} - 2x \right) = \sin -\frac{1}{2} \left( 4x - \pi \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} \\ & = \displaystyle \lim_{x \to \frac{\pi}{4} } \frac{\sin -\frac{1}{2} \left( 4x - \pi \right)}{(4x - \pi)} \\ & = \frac{-\frac{1}{2}}{1} = -\frac{1}{2} \end{align} $
Jadi, hasil limitnya adalah $ -\frac{1}{2} . \, \heartsuit $


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