Soal yang Akan Dibahas
$ \displaystyle \lim_{x \to \infty }
\, \left( x^3\sin \left( \frac{1}{x} \right) + x \right).\left( \frac{1}{x-1} - \frac{1}{x+1} \right) = .... $
A). $ \frac{5}{2} \, $ B). $ 2 \, $ C). $ \frac{3}{2} \, $ D). $ 1 \, $ E). $ \frac{1}{2} $
A). $ \frac{5}{2} \, $ B). $ 2 \, $ C). $ \frac{3}{2} \, $ D). $ 1 \, $ E). $ \frac{1}{2} $
$\spadesuit $ Konsep Dasar
*). Sifat limit trigonometri :
$ \displaystyle \lim_{y \to 0} \frac{\sin ay}{\sin by} = \frac{a}{b} $.
*). Bentuk pecahan : $ a.b = \frac{b}{\frac{1}{a}} $
*). Konsep limit tak hingga aljabar:
$ \displaystyle \lim_{x \to \infty } \frac{cx^2 + ...}{dx^2 + ... } = \frac{c}{d} $.
*). Sifat limit trigonometri :
$ \displaystyle \lim_{y \to 0} \frac{\sin ay}{\sin by} = \frac{a}{b} $.
*). Bentuk pecahan : $ a.b = \frac{b}{\frac{1}{a}} $
*). Konsep limit tak hingga aljabar:
$ \displaystyle \lim_{x \to \infty } \frac{cx^2 + ...}{dx^2 + ... } = \frac{c}{d} $.
$\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 } \, \left( x^3\sin \left( \frac{1}{x} \right) + x \right).\left( \frac{1}{x-1} - \frac{1}{x+1} \right) \\ & = \displaystyle \lim_{x \to \infty } \, x^2\left( x \sin \left( \frac{1}{x} \right) + \frac{1}{x} \right).\left( \frac{(x+1) - (x-1)}{(x-1)(x+1)} \right) \\ & = \displaystyle \lim_{x \to \infty } \, x^2\left( x \sin \left( \frac{1}{x} \right) + \frac{1}{x} \right).\left( \frac{2}{x^2 - 1} \right) \\ & = \displaystyle \lim_{x \to \infty } \, \left( x \sin \left( \frac{1}{x} \right) + \frac{1}{x} \right).\left( \frac{2x^2}{x^2 - 1} \right) \\ & = \displaystyle \lim_{x \to \infty } \, \left( x \sin \left( \frac{1}{x} \right) + \frac{1}{x} \right) . \, \displaystyle \lim_{x \to \infty } \, \left( \frac{2x^2}{x^2 - 1} \right) \\ & = \displaystyle \lim_{x \to \infty } \, \left( \frac{ \sin \left( \frac{1}{x} \right)}{\frac{1}{x}} + \frac{1}{x} \right) . \, \displaystyle \lim_{x \to \infty } \, \left( \frac{2x^2}{x^2 - 1} \right) \\ & = \displaystyle \lim_{y \to 0 } \, \left( \frac{ \sin y}{y} + y \right) . \, \displaystyle \lim_{x \to \infty } \, \left( \frac{2x^2}{x^2 - 1} \right) \\ & = ( 1 + 0 ) . \frac{2}{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 } \, \left( x^3\sin \left( \frac{1}{x} \right) + x \right).\left( \frac{1}{x-1} - \frac{1}{x+1} \right) \\ & = \displaystyle \lim_{x \to \infty } \, x^2\left( x \sin \left( \frac{1}{x} \right) + \frac{1}{x} \right).\left( \frac{(x+1) - (x-1)}{(x-1)(x+1)} \right) \\ & = \displaystyle \lim_{x \to \infty } \, x^2\left( x \sin \left( \frac{1}{x} \right) + \frac{1}{x} \right).\left( \frac{2}{x^2 - 1} \right) \\ & = \displaystyle \lim_{x \to \infty } \, \left( x \sin \left( \frac{1}{x} \right) + \frac{1}{x} \right).\left( \frac{2x^2}{x^2 - 1} \right) \\ & = \displaystyle \lim_{x \to \infty } \, \left( x \sin \left( \frac{1}{x} \right) + \frac{1}{x} \right) . \, \displaystyle \lim_{x \to \infty } \, \left( \frac{2x^2}{x^2 - 1} \right) \\ & = \displaystyle \lim_{x \to \infty } \, \left( \frac{ \sin \left( \frac{1}{x} \right)}{\frac{1}{x}} + \frac{1}{x} \right) . \, \displaystyle \lim_{x \to \infty } \, \left( \frac{2x^2}{x^2 - 1} \right) \\ & = \displaystyle \lim_{y \to 0 } \, \left( \frac{ \sin y}{y} + y \right) . \, \displaystyle \lim_{x \to \infty } \, \left( \frac{2x^2}{x^2 - 1} \right) \\ & = ( 1 + 0 ) . \frac{2}{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.