Soal yang Akan Dibahas
$\displaystyle \lim_{x \to 0} \frac{x(\sqrt{x+1} - 1)}{1 - \cos x} = .... $
A). $ 2 \, $ B). $ 1 \, $ C). $ \frac{1}{2} \, $ D). $ -\frac{1}{2} \, $ E). $ -1 $
A). $ 2 \, $ B). $ 1 \, $ C). $ \frac{1}{2} \, $ D). $ -\frac{1}{2} \, $ E). $ -1 $
$\spadesuit $ Rumus Cepat Limit Trigonometri
$ 1 - \cos px = \frac{1}{2} (px)^2 \, $ sehingga :
$ 1 - cos x = \frac{1}{2} x^2 $.
$ 1 - \cos px = \frac{1}{2} (px)^2 \, $ sehingga :
$ 1 - cos x = \frac{1}{2} x^2 $.
$\clubsuit $ Pembahasan
*). Menyelesaikan Limitnya :
$\begin{align} & \displaystyle \lim_{x \to 0} \frac{x(\sqrt{x+1} - 1)}{1 - \cos x} \\ & = \displaystyle \lim_{x \to 0} \frac{x(\sqrt{x+1} - 1)}{1 - \cos x} . \frac{(\sqrt{x+1} + 1)}{(\sqrt{x+1} + 1)} \\ & = \displaystyle \lim_{x \to 0} \frac{x(\sqrt{x+1} - 1).(\sqrt{x+1} + 1)}{(1 - cos x).(\sqrt{x+1} + 1)} \\ & = \displaystyle \lim_{x \to 0} \frac{x((x + 1) - 1)}{\frac{1}{2}x^2.(\sqrt{x+1} + 1)} \\ & = \displaystyle \lim_{x \to 0} \frac{x.x}{\frac{1}{2}x^2.(\sqrt{x+1} + 1)} \\ & = \displaystyle \lim_{x \to 0} \frac{2}{\sqrt{x+1} + 1} \\ & = \frac{2}{\sqrt{0+1} + 1} = \frac{2}{2} = 1 \end{align} $
Jadi, hasil limitnya adalah $ 1 . \, \heartsuit $
*). Menyelesaikan Limitnya :
$\begin{align} & \displaystyle \lim_{x \to 0} \frac{x(\sqrt{x+1} - 1)}{1 - \cos x} \\ & = \displaystyle \lim_{x \to 0} \frac{x(\sqrt{x+1} - 1)}{1 - \cos x} . \frac{(\sqrt{x+1} + 1)}{(\sqrt{x+1} + 1)} \\ & = \displaystyle \lim_{x \to 0} \frac{x(\sqrt{x+1} - 1).(\sqrt{x+1} + 1)}{(1 - cos x).(\sqrt{x+1} + 1)} \\ & = \displaystyle \lim_{x \to 0} \frac{x((x + 1) - 1)}{\frac{1}{2}x^2.(\sqrt{x+1} + 1)} \\ & = \displaystyle \lim_{x \to 0} \frac{x.x}{\frac{1}{2}x^2.(\sqrt{x+1} + 1)} \\ & = \displaystyle \lim_{x \to 0} \frac{2}{\sqrt{x+1} + 1} \\ & = \frac{2}{\sqrt{0+1} + 1} = \frac{2}{2} = 1 \end{align} $
Jadi, hasil limitnya adalah $ 1 . \, \heartsuit $
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