Kode 248 Pembahasan Limit Trigonometri SBMPTN Matematika IPA tahun 2016

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

$\spadesuit $ Konsep Dasar Limit Trigonometri
*). Sifat Limit Trigonometri :
$ \displaystyle \lim_{ x \to 0 } \frac{\sin ax}{bx} = \frac{a}{b} $
*). Rumus Trigonometri :
$ \cos p x = 1 - 2 \sin^2 \frac{1}{2} (px) \, $ sehingga :
$ 1 - \cos x = 1 - ( 1 - 2 \sin ^2 \frac{1}{2} x ) = 2 . \sin \frac{1}{2} x . \sin \frac{1}{2} x $

$\clubsuit $ Pembahasan
*). Menyelesaikan Limitnya :
$\begin{align} & \displaystyle \lim_{x \to 0} \frac{x(\sqrt{x+1} - 1)}{1 - \cos x} \\ & = \displaystyle \lim_{x \to 0} \frac{x(\sqrt{x+1} - 1)}{1 - \cos x} . \frac{(\sqrt{x+1} + 1)}{(\sqrt{x+1} + 1)} \\ & = \displaystyle \lim_{x \to 0} \frac{x(\sqrt{x+1} - 1).(\sqrt{x+1} + 1)}{(1 - \cos x).(\sqrt{x+1} + 1)} \\ & = \displaystyle \lim_{x \to 0} \frac{x((x + 1) - 1)}{(2 . \sin \frac{1}{2} x . \sin \frac{1}{2} x).(\sqrt{x+1} + 1)} \\ & = \displaystyle \lim_{x \to 0} \frac{x.x}{(2 . \sin \frac{1}{2} x . \sin \frac{1}{2} x).(\sqrt{x+1} + 1)} \\ & = \displaystyle \lim_{x \to 0} \frac{x}{\sin \frac{1}{2} x } . \frac{x}{\sin \frac{1}{2} x } . \frac{1}{2 (\sqrt{x+1} + 1)} \\ & = \frac{1}{\frac{1}{2} } . \frac{1}{ \frac{1}{2} } . \frac{1}{2 (\sqrt{0+1} + 1)} \\ & = 2. 2 . \frac{1}{2 (1 + 1)} \\ & = 2. 2 . \frac{1}{4} = 1 \end{align} $
Jadi, hasil limitnya adalah $ 1 . \, \heartsuit $



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