Soal yang Akan Dibahas
$ \displaystyle \lim_{x \to \infty }
\, x \sec \frac{1}{x} \left(1 - \cos \frac{1}{\sqrt{x}} \right) = .... $
A). $ \frac{1}{2} \, $ B). $ \frac{1}{3} \, $ C). $ \frac{1}{4} \, $ D). $ \frac{1}{5} \, $ E). $ \frac{1}{6} $
A). $ \frac{1}{2} \, $ B). $ \frac{1}{3} \, $ C). $ \frac{1}{4} \, $ D). $ \frac{1}{5} \, $ E). $ \frac{1}{6} $
$\spadesuit $ Konsep Dasar
*). Sifat limit trigonometri :
$ \displaystyle \lim_{y \to 0} \frac{\sin ay}{\sin by} = \frac{a}{b} $.
*). Rumus Trigonometri :
$ \sec A = \frac{1}{\cos A} $
$ \cos A = 1 - 2\sin ^2 \frac{1}{2} A $
$ 1 - \cos A = 2\sin \frac{1}{2} A \sin \frac{1}{2} A $
Sehingga :
$ 1 - \cos y = 2\sin \frac{1}{2}y \sin \frac{1}{2}y $
*). Bentuk pecahan : $ a.b = \frac{b}{\frac{1}{a}} = \frac{b}{\frac{1}{\sqrt{a}} . \frac{1}{\sqrt{a}} } $
*). Sifat limit trigonometri :
$ \displaystyle \lim_{y \to 0} \frac{\sin ay}{\sin by} = \frac{a}{b} $.
*). Rumus Trigonometri :
$ \sec A = \frac{1}{\cos A} $
$ \cos A = 1 - 2\sin ^2 \frac{1}{2} A $
$ 1 - \cos A = 2\sin \frac{1}{2} A \sin \frac{1}{2} A $
Sehingga :
$ 1 - \cos y = 2\sin \frac{1}{2}y \sin \frac{1}{2}y $
*). Bentuk pecahan : $ a.b = \frac{b}{\frac{1}{a}} = \frac{b}{\frac{1}{\sqrt{a}} . \frac{1}{\sqrt{a}} } $
$\clubsuit $ Pembahasan
*). Misalkan $ \frac{1}{\sqrt{x}} = y $, sehingga untuk $ x $ mendekati $ \infty $ maka $ y $ mendekati $0$.
*). Menyelesaikan soal :
$\begin{align} & \displaystyle \lim_{x \to \infty } \, x \sec \frac{1}{x} \left(1 - \cos \frac{1}{\sqrt{x}} \right) \\ & = \displaystyle \lim_{x \to \infty } \, \sec (\frac{1}{\sqrt{x}} . \frac{1}{\sqrt{x}} ) . \frac{\left(1 - \cos \frac{1}{\sqrt{x}} \right)}{\frac{1}{x}} \\ & = \displaystyle \lim_{x \to \infty } \, \sec (\frac{1}{\sqrt{x}} . \frac{1}{\sqrt{x}} ) . \frac{\left(1 - \cos \frac{1}{\sqrt{x}} \right)}{\frac{1}{\sqrt{x}}.\frac{1}{\sqrt{x}}} \\ & = \displaystyle \lim_{y \to 0 } \, \sec y^2 . \frac{\left(1 - \cos y \right)}{y.y} \\ & = \displaystyle \lim_{y \to 0 } \, \frac{1}{\cos y^2} . \frac{2\sin \frac{1}{2}y \sin \frac{1}{2}y}{y.y} \\ & = \displaystyle \lim_{y \to 0 } \, \frac{2}{\cos y^2} . \frac{\sin \frac{1}{2}y}{y} .\frac{ \sin \frac{1}{2}y}{y} \\ & = \frac{2}{\cos 0} . \frac{1}{2} . \frac{1}{2} = \frac{1}{2} \end{align} $
Jadi, hasil limitnya adalah $ \frac{1}{2} . \, \heartsuit $
*). Misalkan $ \frac{1}{\sqrt{x}} = y $, sehingga untuk $ x $ mendekati $ \infty $ maka $ y $ mendekati $0$.
*). Menyelesaikan soal :
$\begin{align} & \displaystyle \lim_{x \to \infty } \, x \sec \frac{1}{x} \left(1 - \cos \frac{1}{\sqrt{x}} \right) \\ & = \displaystyle \lim_{x \to \infty } \, \sec (\frac{1}{\sqrt{x}} . \frac{1}{\sqrt{x}} ) . \frac{\left(1 - \cos \frac{1}{\sqrt{x}} \right)}{\frac{1}{x}} \\ & = \displaystyle \lim_{x \to \infty } \, \sec (\frac{1}{\sqrt{x}} . \frac{1}{\sqrt{x}} ) . \frac{\left(1 - \cos \frac{1}{\sqrt{x}} \right)}{\frac{1}{\sqrt{x}}.\frac{1}{\sqrt{x}}} \\ & = \displaystyle \lim_{y \to 0 } \, \sec y^2 . \frac{\left(1 - \cos y \right)}{y.y} \\ & = \displaystyle \lim_{y \to 0 } \, \frac{1}{\cos y^2} . \frac{2\sin \frac{1}{2}y \sin \frac{1}{2}y}{y.y} \\ & = \displaystyle \lim_{y \to 0 } \, \frac{2}{\cos y^2} . \frac{\sin \frac{1}{2}y}{y} .\frac{ \sin \frac{1}{2}y}{y} \\ & = \frac{2}{\cos 0} . \frac{1}{2} . \frac{1}{2} = \frac{1}{2} \end{align} $
Jadi, hasil limitnya adalah $ \frac{1}{2} . \, \heartsuit $
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