Soal yang Akan Dibahas
$ \displaystyle \lim_{x \to 0}
\, \frac{1 - \sqrt{\cos x }}{2x \sin x} = .... $
A). $ \frac{1}{8} \, $ B). $ \frac{1}{4} \, $ C). $ \frac{\sqrt{2}}{2} \, $ D). $ \frac{\sqrt{3}}{2} \, $ E). $ \frac{1}{2} $
A). $ \frac{1}{8} \, $ B). $ \frac{1}{4} \, $ C). $ \frac{\sqrt{2}}{2} \, $ D). $ \frac{\sqrt{3}}{2} \, $ E). $ \frac{1}{2} $
$\spadesuit $ Konsep Dasar
*). Sifat limit trigonometri :
$ \displaystyle \lim_{x \to 0} \frac{\sin ax}{\sin bx} = \frac{a}{b} \, $ dan $ \, \displaystyle \lim_{x \to 0} \frac{\sin ax}{bx} = \frac{a}{b} $ .
*). Rumus dasar trigonometri :
$ \cos x = 1 - 2\sin ^2 \frac{1}{2} x $
Sehingga : $ 1 - \cos x = 2\sin ^2 \frac{1}{2} x = 2\sin \frac{1}{2} x . \sin \frac{1}{2} x $
*). Merasionalkan :
$ (1 - \sqrt{a})(1 + \sqrt{a}) = 1 - a $
*). Sifat limit trigonometri :
$ \displaystyle \lim_{x \to 0} \frac{\sin ax}{\sin bx} = \frac{a}{b} \, $ dan $ \, \displaystyle \lim_{x \to 0} \frac{\sin ax}{bx} = \frac{a}{b} $ .
*). Rumus dasar trigonometri :
$ \cos x = 1 - 2\sin ^2 \frac{1}{2} x $
Sehingga : $ 1 - \cos x = 2\sin ^2 \frac{1}{2} x = 2\sin \frac{1}{2} x . \sin \frac{1}{2} x $
*). Merasionalkan :
$ (1 - \sqrt{a})(1 + \sqrt{a}) = 1 - a $
$\clubsuit $ Pembahasan
*). Menyelesaikan soal :
$\begin{align} & \displaystyle \lim_{x \to 0} \, \frac{1 - \sqrt{\cos x }}{2x \sin x} \\ & = \displaystyle \lim_{x \to 0} \,\frac{1 - \sqrt{\cos x }}{2x \sin x} \times \frac{1 + \sqrt{\cos x }}{1 + \sqrt{\cos x }} \\ & = \displaystyle \lim_{x \to 0} \,\frac{1 - \cos x }{2x \sin x . (1 + \sqrt{\cos x })} \\ & = \displaystyle \lim_{x \to 0} \,\frac{2\sin \frac{1}{2} x . \sin \frac{1}{2} x}{2x \sin x . (1 + \sqrt{\cos x })} \\ & = \displaystyle \lim_{x \to 0} \,\frac{\sin \frac{1}{2} x . \sin \frac{1}{2} x}{x \sin x . (1 + \sqrt{\cos x })} \\ & = \displaystyle \lim_{x \to 0} \, \frac{\sin \frac{1}{2} x }{x} . \frac{\sin \frac{1}{2} x}{\sin x}. \frac{1}{1 + \sqrt{\cos x }} \\ & = \frac{1}{2} . \frac{1}{2} . \frac{1}{1 + \sqrt{\cos 0 }} \\ & = \frac{1}{2} . \frac{1}{2} . \frac{1}{1 + 1} = \frac{1}{8} \end{align} $
Jadi, hasil limitnya adalah $ \frac{1}{8} . \, \heartsuit $
*). Menyelesaikan soal :
$\begin{align} & \displaystyle \lim_{x \to 0} \, \frac{1 - \sqrt{\cos x }}{2x \sin x} \\ & = \displaystyle \lim_{x \to 0} \,\frac{1 - \sqrt{\cos x }}{2x \sin x} \times \frac{1 + \sqrt{\cos x }}{1 + \sqrt{\cos x }} \\ & = \displaystyle \lim_{x \to 0} \,\frac{1 - \cos x }{2x \sin x . (1 + \sqrt{\cos x })} \\ & = \displaystyle \lim_{x \to 0} \,\frac{2\sin \frac{1}{2} x . \sin \frac{1}{2} x}{2x \sin x . (1 + \sqrt{\cos x })} \\ & = \displaystyle \lim_{x \to 0} \,\frac{\sin \frac{1}{2} x . \sin \frac{1}{2} x}{x \sin x . (1 + \sqrt{\cos x })} \\ & = \displaystyle \lim_{x \to 0} \, \frac{\sin \frac{1}{2} x }{x} . \frac{\sin \frac{1}{2} x}{\sin x}. \frac{1}{1 + \sqrt{\cos x }} \\ & = \frac{1}{2} . \frac{1}{2} . \frac{1}{1 + \sqrt{\cos 0 }} \\ & = \frac{1}{2} . \frac{1}{2} . \frac{1}{1 + 1} = \frac{1}{8} \end{align} $
Jadi, hasil limitnya adalah $ \frac{1}{8} . \, \heartsuit $
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