Soal yang Akan Dibahas
$ \displaystyle \lim_{x \to \infty }
\, \frac{2x^2 \tan \left( \frac{1}{x} \right) - x \sin \left( \frac{1}{x} \right) + \frac{1}{x}}{x \cos \left( \frac{2}{x} \right)} = .... $
A). $ 2 \, $ B). $ 1 \, $ C). $ 0 \, $ D). $ -1 \, $ E). $ -2 $
A). $ 2 \, $ B). $ 1 \, $ C). $ 0 \, $ D). $ -1 \, $ E). $ -2 $
$\spadesuit $ Konsep Dasar
*). Sifat limit trigonometri :
$ \displaystyle \lim_{y \to 0} \frac{\tan y}{y} = 1 $.
*). Sifat limit trigonometri :
$ \displaystyle \lim_{y \to 0} \frac{\tan y}{y} = 1 $.
$\clubsuit $ Pembahasan
*). Misalkan $ \frac{1}{x} = y $, sehingga untuk $ x $ mendekati $ \infty $ maka $ y $ mendekati $0$.
*). Menyelesaikan soal :
$\begin{align} & \displaystyle \lim_{x \to \infty } \, \frac{2x^2 \tan \left( \frac{1}{x} \right) - x \sin \left( \frac{1}{x} \right) + \frac{1}{x}}{x \cos \left( \frac{2}{x} \right)} \times \frac{\frac{1}{x}}{\frac{1}{x}} \\ & = \displaystyle \lim_{x \to \infty } \, \frac{2x \tan \left( \frac{1}{x} \right) - \sin \left( \frac{1}{x} \right) + \frac{1}{x}.\frac{1}{x}}{ \cos \left( \frac{2}{x} \right)} \\ & = \displaystyle \lim_{x \to \infty } \, \left( 2x \tan \left( \frac{1}{x} \right) - \sin \left( \frac{1}{x} \right) + \frac{1}{x}.\frac{1}{x} \right) \times \frac{1}{ \cos \left( \frac{2}{x} \right)} \\ & = \displaystyle \lim_{x \to \infty } \, \left( 2 \frac{ \tan \left( \frac{1}{x} \right)}{\frac{1}{x}} - \sin \left( \frac{1}{x} \right) + \frac{1}{x}.\frac{1}{x} \right) \times \frac{1}{ \cos \left( \frac{2}{x} \right)} \\ & = \displaystyle \lim_{y \to 0 } \, \left( 2 \frac{ \tan y}{y} - \sin y + y^2 \right) \times \frac{1}{ \cos 2y} \\ & = \left( 2 . 1 - \sin 0 + 0^2 \right) \times \frac{1}{ \cos 0} \\ & = \left( 2 \right) \times \frac{1}{1} = 2 \end{align} $
Jadi, hasil limitnya adalah $ 2 . \, \heartsuit $
*). Misalkan $ \frac{1}{x} = y $, sehingga untuk $ x $ mendekati $ \infty $ maka $ y $ mendekati $0$.
*). Menyelesaikan soal :
$\begin{align} & \displaystyle \lim_{x \to \infty } \, \frac{2x^2 \tan \left( \frac{1}{x} \right) - x \sin \left( \frac{1}{x} \right) + \frac{1}{x}}{x \cos \left( \frac{2}{x} \right)} \times \frac{\frac{1}{x}}{\frac{1}{x}} \\ & = \displaystyle \lim_{x \to \infty } \, \frac{2x \tan \left( \frac{1}{x} \right) - \sin \left( \frac{1}{x} \right) + \frac{1}{x}.\frac{1}{x}}{ \cos \left( \frac{2}{x} \right)} \\ & = \displaystyle \lim_{x \to \infty } \, \left( 2x \tan \left( \frac{1}{x} \right) - \sin \left( \frac{1}{x} \right) + \frac{1}{x}.\frac{1}{x} \right) \times \frac{1}{ \cos \left( \frac{2}{x} \right)} \\ & = \displaystyle \lim_{x \to \infty } \, \left( 2 \frac{ \tan \left( \frac{1}{x} \right)}{\frac{1}{x}} - \sin \left( \frac{1}{x} \right) + \frac{1}{x}.\frac{1}{x} \right) \times \frac{1}{ \cos \left( \frac{2}{x} \right)} \\ & = \displaystyle \lim_{y \to 0 } \, \left( 2 \frac{ \tan y}{y} - \sin y + y^2 \right) \times \frac{1}{ \cos 2y} \\ & = \left( 2 . 1 - \sin 0 + 0^2 \right) \times \frac{1}{ \cos 0} \\ & = \left( 2 \right) \times \frac{1}{1} = 2 \end{align} $
Jadi, hasil limitnya adalah $ 2 . \, \heartsuit $
Tidak ada komentar:
Posting Komentar
Catatan: Hanya anggota dari blog ini yang dapat mengirim komentar.