Kode 246 Pembahasan Limit Trigonometri SBMPTN Matematika IPA tahun 2016

Soal yang Akan Dibahas
$\displaystyle \lim_{x \to 0} \frac{x^2\sin (x) - \left(\frac{1}{2}\right)\sin (x) \sqrt{x}}{x^\frac{3}{2}} = .... $
A). $ - \infty \, $ B). $ -\frac{7}{2} \, $ C). $ -\frac{5}{2} \, $ D). $ -\frac{3}{2} \, $ E). $ -\frac{1}{2} $

$\spadesuit $ Konsep Dasar Limit Trigonometri
*). Sifat Limit Trigonometri :
$ \displaystyle \lim_{ x \to 0 } \frac{\sin x}{x} = 1 $
*). Sifat-sifat Eksponen :
$ a^{m+n} = a^m . a^n $
$ a^\frac{1}{2} = \sqrt{a} $

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



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