Soal yang Akan Dibahas
$ \displaystyle \lim_{x \to 0}
\frac{-\tan x \sec x + \sin x}{x (\cos x - 1)} = .... $
A). $ -1 \, $ B). $ 0 \, $ C). $ 1 \, $ D). $ 2 \, $ E). $ 3 $
A). $ -1 \, $ B). $ 0 \, $ C). $ 1 \, $ D). $ 2 \, $ E). $ 3 $
$\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 :
$ \tan x = \frac{\sin x}{\cos x} \, $ dan $ \sec x = \frac{1}{\cos x} $
$ \sin ^2 x + \cos ^2 x = 1 \rightarrow 1 - \cos ^2 x = \sin ^2 x $
$ \cos x = 1 - 2\sin ^2 \frac{1}{2} x $
*). Bentuk pecahan : $ \frac{\frac{a}{b}}{c} = \frac{a}{bc} $
*). 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 :
$ \tan x = \frac{\sin x}{\cos x} \, $ dan $ \sec x = \frac{1}{\cos x} $
$ \sin ^2 x + \cos ^2 x = 1 \rightarrow 1 - \cos ^2 x = \sin ^2 x $
$ \cos x = 1 - 2\sin ^2 \frac{1}{2} x $
*). Bentuk pecahan : $ \frac{\frac{a}{b}}{c} = \frac{a}{bc} $
$\clubsuit $ Pembahasan
*). Menyederhanakan soal :
$\begin{align} -\tan x \sec x + \sin x & = -\frac{\sin x}{\cos x} . \frac{1}{\cos x} + \sin x \\ & = -\frac{\sin x}{\cos ^2 x} + \sin x \\ & = \frac{-\sin x + \sin x \cos ^2 x }{\cos ^2 x} \\ & = \frac{-\sin x ( 1 - \cos ^2 x ) }{\cos ^2 x} \\ & = \frac{-\sin x .\sin ^2 x }{\cos ^2 x} \\ & = \frac{-\sin ^3 x }{\cos ^2 x} \\ \cos x - 1 & = (1 - 2\sin ^2 \frac{1}{2} x ) - 1 \\ & = - 2\sin ^2 \frac{1}{2} x \end{align} $
*). Menyelesaikan soal :
$\begin{align} & \displaystyle \lim_{x \to 0} \frac{-\tan x \sec x + \sin x}{x (\cos x - 1)} \\ & = \displaystyle \lim_{x \to 0} \frac{\frac{-\sin ^3 x }{\cos ^2 x} }{x(- 2\sin ^2 \frac{1}{2} x)} \\ & = \displaystyle \lim_{x \to 0} \frac{\sin ^3 x }{ 2x\sin ^2 \frac{1}{2} x . \cos ^2 x} \\ & = \displaystyle \lim_{x \to 0} \left( \frac{\sin x }{ 2x}.\frac{\sin x }{ \sin \frac{1}{2} x }. \frac{\sin x }{ \sin \frac{1}{2} x }.\frac{1}{ \cos ^2 x} \right) \\ & = \frac{1}{2} . \frac{1}{\frac{1}{2}}.\frac{1}{\frac{1}{2}}. \frac{1}{ \cos ^2 0} \\ & = \frac{1}{2} .2.2. \frac{1}{ 1} = 2 \end{align} $
Jadi, hasil limitnya adalah $ 2 . \, \heartsuit $
*). Menyederhanakan soal :
$\begin{align} -\tan x \sec x + \sin x & = -\frac{\sin x}{\cos x} . \frac{1}{\cos x} + \sin x \\ & = -\frac{\sin x}{\cos ^2 x} + \sin x \\ & = \frac{-\sin x + \sin x \cos ^2 x }{\cos ^2 x} \\ & = \frac{-\sin x ( 1 - \cos ^2 x ) }{\cos ^2 x} \\ & = \frac{-\sin x .\sin ^2 x }{\cos ^2 x} \\ & = \frac{-\sin ^3 x }{\cos ^2 x} \\ \cos x - 1 & = (1 - 2\sin ^2 \frac{1}{2} x ) - 1 \\ & = - 2\sin ^2 \frac{1}{2} x \end{align} $
*). Menyelesaikan soal :
$\begin{align} & \displaystyle \lim_{x \to 0} \frac{-\tan x \sec x + \sin x}{x (\cos x - 1)} \\ & = \displaystyle \lim_{x \to 0} \frac{\frac{-\sin ^3 x }{\cos ^2 x} }{x(- 2\sin ^2 \frac{1}{2} x)} \\ & = \displaystyle \lim_{x \to 0} \frac{\sin ^3 x }{ 2x\sin ^2 \frac{1}{2} x . \cos ^2 x} \\ & = \displaystyle \lim_{x \to 0} \left( \frac{\sin x }{ 2x}.\frac{\sin x }{ \sin \frac{1}{2} x }. \frac{\sin x }{ \sin \frac{1}{2} x }.\frac{1}{ \cos ^2 x} \right) \\ & = \frac{1}{2} . \frac{1}{\frac{1}{2}}.\frac{1}{\frac{1}{2}}. \frac{1}{ \cos ^2 0} \\ & = \frac{1}{2} .2.2. \frac{1}{ 1} = 2 \end{align} $
Jadi, hasil limitnya adalah $ 2 . \, \heartsuit $
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