Pembahasan Limit Trigonometri SBMPTN 2017 Matematika IPA kode 165

Soal yang Akan Dibahas
$ \displaystyle \lim_{x \to -3} \frac{\tan (x+3)}{(x^2-2x-15)\sin \left(\frac{\pi}{2}x\right)} = .... $
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 k} \frac{\tan f(x)}{f(x)} = 1 \, $ dengan syarat $ f(k) = 0 $.
*). Trigonometri :
$ \sin (-x) = -\sin x $.
Sehingga $ \sin (-270^\circ ) = -\sin 270^\circ = - (-1) = 1 $.

$\clubsuit $ Pembahasan
*). Menyelesaikan soal :
$\begin{align} & \displaystyle \lim_{x \to -3} \frac{\tan (x+3)}{(x^2-2x-15)\sin \left(\frac{\pi}{2}x\right)} \\ & = \displaystyle \lim_{x \to -3} \frac{\tan (x+3)}{(x+3)(x-5)\sin \left(\frac{\pi}{2}x\right)} \\ & = \displaystyle \lim_{x \to -3} \frac{\tan (x+3)}{(x+3)} \times \displaystyle \lim_{x \to -3} \frac{1}{(x-5)\sin \left(\frac{\pi}{2}x\right)} \\ & = 1 \times \frac{1}{(-3-5)\sin \left(\frac{\pi}{2}\times -3 \right)} \\ & = \frac{1}{-8 . \sin (-270^\circ )} \\ & = \frac{1}{-8 . 1} = \frac{1}{-8} = -\frac{1}{8} \end{align} $
Jadi, hasil limitnya adalah $ -\frac{1}{8} . \, \heartsuit $

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