Soal yang Akan Dibahas
$\displaystyle \lim_{h \to 0} \frac{\tan (-x + h) - \tan (-x-h)}{h\sqrt{4-h^2}} = .... $
A). $ \sec ^2 x \, $ B). $ 2\sec ^2 x \, $ C). $ 4\sec ^2 x \, $ D). $ \sec x \, $ E). $2\sec x $
A). $ \sec ^2 x \, $ B). $ 2\sec ^2 x \, $ C). $ 4\sec ^2 x \, $ D). $ \sec x \, $ E). $2\sec x $
$\spadesuit $ Konsep Dasar Limit Trigonometri
*). Konsep Trigonometri :
$ \tan A = \frac{\sin A}{\cos A} \, $ dan $ \sec A = \frac{1}{\cos A} $
$ \sin (A-B) = \sin A \cos B - \cos A \sin B $
$ \cos (-A) = \cos A $
*). Sifat Limit Trigonometri :
$ \displaystyle \lim_{ x \to 0 } \frac{\sin ax}{bx} = \frac{a}{b} $
*). Konsep Trigonometri :
$ \tan A = \frac{\sin A}{\cos A} \, $ dan $ \sec A = \frac{1}{\cos A} $
$ \sin (A-B) = \sin A \cos B - \cos A \sin B $
$ \cos (-A) = \cos A $
*). Sifat Limit Trigonometri :
$ \displaystyle \lim_{ x \to 0 } \frac{\sin ax}{bx} = \frac{a}{b} $
$\clubsuit $ Pembahasan
*). Mengubah bentuk dengan $ \sin (A-B) = \sin A \cos B - \cos A \sin B $
$\begin{align} & \tan ( - x + h ) - \tan (-x - h) \\ & = \frac{\sin (-x+h)}{\cos (-x+h)} - \frac{\sin (-x-h)}{\cos (-x-h)} \\ & = \frac{\sin (-x+h) \cos (-x-h)}{\cos (-x+h) \cos (-x-h)} - \frac{\cos (-x + h)\sin (-x-h)}{\cos (-x + h)\cos (-x-h)} \\ & = \frac{\sin (-x+h) \cos (-x-h) - \cos (-x + h)\sin (-x-h) }{\cos (-x+h) \cos (-x-h)} \\ & = \frac{\sin [(-x+h) - (-x-h) ] }{\cos (-x+h) \cos (-x-h)} \\ & = \frac{\sin 2h }{\cos (-x+h) \cos (-x-h)} \end{align} $
*). Menyelesaikan soal :
$\begin{align} & \displaystyle \lim_{h \to 0} \frac{\tan (-x + h) - \tan (-x-h)}{h\sqrt{4-h^2}} \\ & = \displaystyle \lim_{h \to 0} \frac{\frac{\sin 2h }{\cos (-x+h) \cos (-x-h)} }{h\sqrt{4-h^2}} \\ & = \displaystyle \lim_{h \to 0} \frac{\sin 2h }{h} . \frac{1}{\sqrt{4-h^2} \cos (-x+h) \cos (-x-h)} \\ & = \frac{2 }{1} . \frac{1}{\sqrt{4-0^2} \cos (-x+0) \cos (-x-0)} \\ & = 2. \frac{1}{2\cos (-x ) \cos (-x )} \\ & = \frac{1}{ \cos x \cos x} \\ & = \frac{1}{ \cos ^2 x } \\ & = \sec ^2 x \end{align} $
Jadi, hasil limitnya adalah $ \sec ^2 x . \, \heartsuit $
*). Mengubah bentuk dengan $ \sin (A-B) = \sin A \cos B - \cos A \sin B $
$\begin{align} & \tan ( - x + h ) - \tan (-x - h) \\ & = \frac{\sin (-x+h)}{\cos (-x+h)} - \frac{\sin (-x-h)}{\cos (-x-h)} \\ & = \frac{\sin (-x+h) \cos (-x-h)}{\cos (-x+h) \cos (-x-h)} - \frac{\cos (-x + h)\sin (-x-h)}{\cos (-x + h)\cos (-x-h)} \\ & = \frac{\sin (-x+h) \cos (-x-h) - \cos (-x + h)\sin (-x-h) }{\cos (-x+h) \cos (-x-h)} \\ & = \frac{\sin [(-x+h) - (-x-h) ] }{\cos (-x+h) \cos (-x-h)} \\ & = \frac{\sin 2h }{\cos (-x+h) \cos (-x-h)} \end{align} $
*). Menyelesaikan soal :
$\begin{align} & \displaystyle \lim_{h \to 0} \frac{\tan (-x + h) - \tan (-x-h)}{h\sqrt{4-h^2}} \\ & = \displaystyle \lim_{h \to 0} \frac{\frac{\sin 2h }{\cos (-x+h) \cos (-x-h)} }{h\sqrt{4-h^2}} \\ & = \displaystyle \lim_{h \to 0} \frac{\sin 2h }{h} . \frac{1}{\sqrt{4-h^2} \cos (-x+h) \cos (-x-h)} \\ & = \frac{2 }{1} . \frac{1}{\sqrt{4-0^2} \cos (-x+0) \cos (-x-0)} \\ & = 2. \frac{1}{2\cos (-x ) \cos (-x )} \\ & = \frac{1}{ \cos x \cos x} \\ & = \frac{1}{ \cos ^2 x } \\ & = \sec ^2 x \end{align} $
Jadi, hasil limitnya adalah $ \sec ^2 x . \, \heartsuit $
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