Soal yang Akan Dibahas
$ \displaystyle \lim_{x \to 0} \,
\frac{\sec x + \cos x - 2}{x^2 \sin ^2 x} = .... $
A). $ -\frac{1}{8} \, $ B). $ -\frac{1}{4} \, $ C). $ 0 \, $ D). $ \frac{1}{4} \, $ E). $ \frac{1}{8} $
A). $ -\frac{1}{8} \, $ B). $ -\frac{1}{4} \, $ C). $ 0 \, $ D). $ \frac{1}{4} \, $ E). $ \frac{1}{8} $
$\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 :
$ \sec x = \frac{1}{ \cos x} $ dan $ \cos x = 1 - 2\sin ^2 \frac{1}{2} x $
Sehingga : $ 1 - \cos x = 2\sin ^2 \frac{1}{2} x $
*). 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 :
$ \sec x = \frac{1}{ \cos x} $ dan $ \cos x = 1 - 2\sin ^2 \frac{1}{2} x $
Sehingga : $ 1 - \cos x = 2\sin ^2 \frac{1}{2} x $
$\clubsuit $ Pembahasan
*). Menyelesaikan soal :
$\begin{align} & \displaystyle \lim_{x \to 0} \, \frac{\sec x + \cos x - 2}{x^2 \sin x} \\ & = \displaystyle \lim_{x \to 0} \, \frac{\frac{1}{\cos x } + \cos x - 2}{x^2 \sin ^2 x} \\ & = \displaystyle \lim_{x \to 0} \, \frac{\frac{1 + \cos ^2 x - 2\cos x }{\cos x } }{x^2 \sin ^2 x} \\ & = \displaystyle \lim_{x \to 0} \, \frac{ 1 + \cos ^2 x - 2\cos x }{x^2 \sin ^2 x . \cos x } \\ & = \displaystyle \lim_{x \to 0} \, \frac{ ( 1 - \cos x )^2 }{x^2 \sin ^2 x . \cos x } \\ & = \displaystyle \lim_{x \to 0} \, \frac{ ( 2 \sin ^2 \frac{1}{2} x )^2 }{x^2 \sin ^2 x . \cos x } \\ & = \displaystyle \lim_{x \to 0} \, \frac{ 4 \sin ^4 \frac{1}{2} x }{x^2 \sin ^2 x . \cos x } \\ & = \displaystyle \lim_{x \to 0} \, \frac{\sin \frac{1}{2} x }{ x} . \frac{\sin \frac{1}{2} x }{ x} . \frac{\sin \frac{1}{2} x }{ \sin x} .\frac{\sin \frac{1}{2} x }{ \sin x} . \frac{ 4 }{ \cos x } \\ & = \frac{1}{2} . \frac{1}{2} . \frac{1}{2} . \frac{1}{2} . \frac{ 4 }{ \cos 0 } \\ & = \frac{1}{16} . \frac{ 4 }{ 1 } = \frac{1}{4} \end{align} $
Jadi, hasil limitnya adalah $ \frac{1}{4} . \, \heartsuit $
*). Menyelesaikan soal :
$\begin{align} & \displaystyle \lim_{x \to 0} \, \frac{\sec x + \cos x - 2}{x^2 \sin x} \\ & = \displaystyle \lim_{x \to 0} \, \frac{\frac{1}{\cos x } + \cos x - 2}{x^2 \sin ^2 x} \\ & = \displaystyle \lim_{x \to 0} \, \frac{\frac{1 + \cos ^2 x - 2\cos x }{\cos x } }{x^2 \sin ^2 x} \\ & = \displaystyle \lim_{x \to 0} \, \frac{ 1 + \cos ^2 x - 2\cos x }{x^2 \sin ^2 x . \cos x } \\ & = \displaystyle \lim_{x \to 0} \, \frac{ ( 1 - \cos x )^2 }{x^2 \sin ^2 x . \cos x } \\ & = \displaystyle \lim_{x \to 0} \, \frac{ ( 2 \sin ^2 \frac{1}{2} x )^2 }{x^2 \sin ^2 x . \cos x } \\ & = \displaystyle \lim_{x \to 0} \, \frac{ 4 \sin ^4 \frac{1}{2} x }{x^2 \sin ^2 x . \cos x } \\ & = \displaystyle \lim_{x \to 0} \, \frac{\sin \frac{1}{2} x }{ x} . \frac{\sin \frac{1}{2} x }{ x} . \frac{\sin \frac{1}{2} x }{ \sin x} .\frac{\sin \frac{1}{2} x }{ \sin x} . \frac{ 4 }{ \cos x } \\ & = \frac{1}{2} . \frac{1}{2} . \frac{1}{2} . \frac{1}{2} . \frac{ 4 }{ \cos 0 } \\ & = \frac{1}{16} . \frac{ 4 }{ 1 } = \frac{1}{4} \end{align} $
Jadi, hasil limitnya adalah $ \frac{1}{4} . \, \heartsuit $
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