Kode 247 Pembahasan Limit Trigonometri SBMPTN Matematika IPA tahun 2016

Soal yang Akan Dibahas
$\displaystyle \lim_{x \to 0} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3} = .... $
A). $ -\frac{1}{2} \, $ B). $ -\frac{1}{4} \, $ C). $ 0 \, $ D). $ \frac{1}{4} \, $ E). $ \frac{1}{2} $

$\spadesuit $ Konsep Dasar Limit Trigonometri
*). Sifat Limit Trigonometri :
$ \displaystyle \lim_{ x \to 0 } \frac{\sin ax}{bx} = \frac{a}{b} $
*). Rumus trigonometri :
$\tan x = \frac{\sin x}{\cos x} $ dan $ \cos px = 1 - 2 \sin ^2 \frac{1}{2} p x $
Sehingga :
$ \begin{align} 1 - \cos x & = 1 - (1 -2\sin ^2 \frac{1}{2} x ) \\ & = 2 \sin \frac{1}{2}x \sin \frac{1}{2} x \\ \tan x - \sin x & = \frac{\sin x}{\cos x} - \sin x \\ & = \frac{\sin x}{\cos x} - \frac{\sin x \cos x}{\cos x} \\ & = \frac{\sin x - \sin x \cos x }{\cos x} \\ & = \frac{\sin x ( 1 - \cos x ) }{\cos x} \\ & = \frac{\sin x . 2 \sin \frac{1}{2}x . \sin \frac{1}{2} x }{\cos x} \end{align} $
*). Sifat-sifat Eksponen :
$ (\sqrt{a} - \sqrt{b}) (\sqrt{a} + \sqrt{b}) = a - b $

$\clubsuit $ Pembahasan
*). Menyelesaikan Limitnya :
$\begin{align} & \displaystyle \lim_{x \to 0} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3} \\ & = \displaystyle \lim_{x \to 0} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3} . \frac{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}}{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}} \\ & = \displaystyle \lim_{x \to 0} \frac{(1 + \tan x) -(1 + \sin x)}{x^3. (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})} \\ & = \displaystyle \lim_{x \to 0} \frac{\tan x - \sin x }{x^3. (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})} \\ & = \displaystyle \lim_{x \to 0} \frac{\frac{\sin x . 2 \sin \frac{1}{2}x . \sin \frac{1}{2} x }{\cos x}}{x^3. (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})} \\ & = \displaystyle \lim_{x \to 0} \frac{\sin x . 2 \sin \frac{1}{2}x . \sin \frac{1}{2} x }{x^3. \cos x (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})} \\ & = \displaystyle \lim_{x \to 0} \frac{\sin x}{x} . \frac{\sin \frac{1}{2}x}{x} . \frac{\sin \frac{1}{2}x}{x} . \frac{2 }{ \cos x (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})} \\ & = 1 . \frac{1}{2} . \frac{1}{2} . \frac{2 }{ \cos 0 . (\sqrt{1 + \tan 0} + \sqrt{1 + \sin 0})} \\ & = \frac{1}{4} . \frac{2 }{ 1 . (\sqrt{1 + 0} + \sqrt{1 + 0})} \\ & = \frac{1}{4} . \frac{2 }{ 2} \\ & = \frac{1}{4} \end{align} $
Jadi, hasil limitnya adalah $ \frac{1}{4}. \, \heartsuit $



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