Soal yang Akan Dibahas
$ \displaystyle \lim_{x \to \infty }
\, \frac{\csc ^2 \left( \frac{2}{x} \right) - x^3 \sin \left( \frac{4}{x} \right)}{x^2} = .... $
A). $ -\frac{23}{4} \, $ B). $ -\frac{21}{4} \, $ C). $ -\frac{19}{4} \, $ D). $ -\frac{17}{4} \, $ E). $ -\frac{15}{4} \, $
A). $ -\frac{23}{4} \, $ B). $ -\frac{21}{4} \, $ C). $ -\frac{19}{4} \, $ D). $ -\frac{17}{4} \, $ E). $ -\frac{15}{4} \, $
$\spadesuit $ Konsep Dasar
*). Sifat limit trigonometri :
$ \displaystyle \lim_{y \to 0} \frac{ ay}{\sin by} = \frac{a}{b} $ dan $ \displaystyle \lim_{y \to 0} \frac{ \sin ay}{ by} = \frac{a}{b} $
*). Rumus dasar trigonometri :
$ \csc A = \frac{1}{\sin A} $
*). Sifat limit trigonometri :
$ \displaystyle \lim_{y \to 0} \frac{ ay}{\sin by} = \frac{a}{b} $ dan $ \displaystyle \lim_{y \to 0} \frac{ \sin ay}{ by} = \frac{a}{b} $
*). Rumus dasar trigonometri :
$ \csc A = \frac{1}{\sin A} $
$\clubsuit $ Pembahasan
*). Misalkan $ \frac{1}{x} = y \rightarrow x = \frac{1}{y}$, sehingga untuk $ x $ mendekati $ \infty $ maka $ y $ mendekati $0$.
*). Menyelesaikan soal :
$\begin{align} & \displaystyle \lim_{x \to \infty } \, \frac{\csc ^2 \left( \frac{2}{x} \right) - x^3 \sin \left( \frac{4}{x} \right)}{x^2} \\ & = \displaystyle \lim_{y \to 0 } \, \frac{\csc ^2 2y - (\frac{1}{y})^3 \sin 4y}{(\frac{1}{y})^2} \\ & = \displaystyle \lim_{y \to 0 } \, \frac{\csc ^2 2y - \frac{1}{y^3}. \sin 4y}{\frac{1}{y^2} } \times \frac{y^2}{y^2} \\ & = \displaystyle \lim_{y \to 0 } \, \frac{y^2\csc ^2 2y - \frac{1}{y} \sin 4y}{1} \\ & = \displaystyle \lim_{y \to 0 } \, \left( y^2 . \frac{1}{\sin ^2 2y} - \frac{\sin 4y}{y} \right) \\ & = \displaystyle \lim_{y \to 0 } \, \left( \frac{y^2}{\sin ^2 2y} - \frac{\sin 4y}{y} \right) \\ & = \displaystyle \lim_{y \to 0 } \, \left( \frac{y }{\sin 2y}.\frac{y }{\sin 2y} - \frac{\sin 4y}{y} \right) \\ & = \frac{1}{2}.\frac{1}{2} - 4 = \frac{1}{4} - 4 \\ & = \frac{1}{4} - \frac{16}{4} = -\frac{15}{4} \end{align} $
Jadi, hasil limitnya adalah $ -\frac{15}{4} . \, \heartsuit $
*). Misalkan $ \frac{1}{x} = y \rightarrow x = \frac{1}{y}$, sehingga untuk $ x $ mendekati $ \infty $ maka $ y $ mendekati $0$.
*). Menyelesaikan soal :
$\begin{align} & \displaystyle \lim_{x \to \infty } \, \frac{\csc ^2 \left( \frac{2}{x} \right) - x^3 \sin \left( \frac{4}{x} \right)}{x^2} \\ & = \displaystyle \lim_{y \to 0 } \, \frac{\csc ^2 2y - (\frac{1}{y})^3 \sin 4y}{(\frac{1}{y})^2} \\ & = \displaystyle \lim_{y \to 0 } \, \frac{\csc ^2 2y - \frac{1}{y^3}. \sin 4y}{\frac{1}{y^2} } \times \frac{y^2}{y^2} \\ & = \displaystyle \lim_{y \to 0 } \, \frac{y^2\csc ^2 2y - \frac{1}{y} \sin 4y}{1} \\ & = \displaystyle \lim_{y \to 0 } \, \left( y^2 . \frac{1}{\sin ^2 2y} - \frac{\sin 4y}{y} \right) \\ & = \displaystyle \lim_{y \to 0 } \, \left( \frac{y^2}{\sin ^2 2y} - \frac{\sin 4y}{y} \right) \\ & = \displaystyle \lim_{y \to 0 } \, \left( \frac{y }{\sin 2y}.\frac{y }{\sin 2y} - \frac{\sin 4y}{y} \right) \\ & = \frac{1}{2}.\frac{1}{2} - 4 = \frac{1}{4} - 4 \\ & = \frac{1}{4} - \frac{16}{4} = -\frac{15}{4} \end{align} $
Jadi, hasil limitnya adalah $ -\frac{15}{4} . \, \heartsuit $
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