Soal yang Akan Dibahas
$\displaystyle \lim_{x \to 0} \frac{\sqrt{x^2 + 1} - 1}{\sqrt{3x^5 + 4 \sin ^4 x}} = .... $
A). $ 0 \, $ B). $ \frac{1}{4} \, $ C). $ \frac{1}{\sqrt{7}} \, $ D). $ \frac{1}{2} \, $ E). $ \frac{1}{\sqrt{3}} $
A). $ 0 \, $ B). $ \frac{1}{4} \, $ C). $ \frac{1}{\sqrt{7}} \, $ D). $ \frac{1}{2} \, $ E). $ \frac{1}{\sqrt{3}} $
$\spadesuit $ Konsep Dasar Limit Trigonometri
*). Sifat Limit Trigonometri :
$ \displaystyle \lim_{ x \to 0 } \frac{\sin x}{x} = 1 $ dan $ \displaystyle \lim_{ x \to 0 } \frac{\sin ^n x}{x^n} = 1 $
*). Sifat eksponen :
$ \sqrt{a.b} = \sqrt{a} . \sqrt{b} $
*). Sifat Limit Trigonometri :
$ \displaystyle \lim_{ x \to 0 } \frac{\sin x}{x} = 1 $ dan $ \displaystyle \lim_{ x \to 0 } \frac{\sin ^n x}{x^n} = 1 $
*). Sifat eksponen :
$ \sqrt{a.b} = \sqrt{a} . \sqrt{b} $
$\clubsuit $ Pembahasan
*). Menyelesaikan Limitnya :
$\begin{align} & \displaystyle \lim_{x \to 0} \frac{\sqrt{x^2 + 1} - 1}{\sqrt{3x^5 + 4 \sin ^4 x}} \\ & = \displaystyle \lim_{x \to 0} \frac{\sqrt{x^2 + 1} - 1}{\sqrt{3x^5 + 4 \sin ^4 x}} . \frac{\sqrt{x^2 + 1} + 1}{\sqrt{x^2 + 1} + 1} \\ & = \displaystyle \lim_{x \to 0} \frac{(x^2 + 1) - 1}{\sqrt{3x^5 + 4 \sin ^4 x} . (\sqrt{x^2 + 1} + 1)} \\ & = \displaystyle \lim_{x \to 0} \frac{x^2}{\sqrt{x^4(3x + \frac{4 \sin ^4 x}{x^4})} . (\sqrt{x^2 + 1} + 1)} \\ & = \displaystyle \lim_{x \to 0} \frac{x^2}{\sqrt{x^4}\sqrt{3x + \frac{4 \sin ^4 x}{x^4}} . (\sqrt{x^2 + 1} + 1)} \\ & = \displaystyle \lim_{x \to 0} \frac{x^2}{x^2\sqrt{3x + \frac{4 \sin ^4 x}{x^4}} . (\sqrt{x^2 + 1} + 1)} \\ & = \displaystyle \lim_{x \to 0} \frac{1}{\sqrt{3x + 4.\frac{ \sin ^4 x}{x^4}} . (\sqrt{x^2 + 1} + 1)} \\ & = \frac{1}{\sqrt{3.0 + 4.1} . (\sqrt{0^2 + 1} + 1)} \\ & = \frac{1}{\sqrt{0 + 4} . (\sqrt{0 + 1} + 1)} \\ & = \frac{1}{\sqrt{4} . (\sqrt{ 1} + 1)} = \frac{1}{2 . (2)} = \frac{1}{4} \end{align} $
Jadi, hasil limitnya adalah $ \frac{1}{4}. \, \heartsuit $
*). Menyelesaikan Limitnya :
$\begin{align} & \displaystyle \lim_{x \to 0} \frac{\sqrt{x^2 + 1} - 1}{\sqrt{3x^5 + 4 \sin ^4 x}} \\ & = \displaystyle \lim_{x \to 0} \frac{\sqrt{x^2 + 1} - 1}{\sqrt{3x^5 + 4 \sin ^4 x}} . \frac{\sqrt{x^2 + 1} + 1}{\sqrt{x^2 + 1} + 1} \\ & = \displaystyle \lim_{x \to 0} \frac{(x^2 + 1) - 1}{\sqrt{3x^5 + 4 \sin ^4 x} . (\sqrt{x^2 + 1} + 1)} \\ & = \displaystyle \lim_{x \to 0} \frac{x^2}{\sqrt{x^4(3x + \frac{4 \sin ^4 x}{x^4})} . (\sqrt{x^2 + 1} + 1)} \\ & = \displaystyle \lim_{x \to 0} \frac{x^2}{\sqrt{x^4}\sqrt{3x + \frac{4 \sin ^4 x}{x^4}} . (\sqrt{x^2 + 1} + 1)} \\ & = \displaystyle \lim_{x \to 0} \frac{x^2}{x^2\sqrt{3x + \frac{4 \sin ^4 x}{x^4}} . (\sqrt{x^2 + 1} + 1)} \\ & = \displaystyle \lim_{x \to 0} \frac{1}{\sqrt{3x + 4.\frac{ \sin ^4 x}{x^4}} . (\sqrt{x^2 + 1} + 1)} \\ & = \frac{1}{\sqrt{3.0 + 4.1} . (\sqrt{0^2 + 1} + 1)} \\ & = \frac{1}{\sqrt{0 + 4} . (\sqrt{0 + 1} + 1)} \\ & = \frac{1}{\sqrt{4} . (\sqrt{ 1} + 1)} = \frac{1}{2 . (2)} = \frac{1}{4} \end{align} $
Jadi, hasil limitnya adalah $ \frac{1}{4}. \, \heartsuit $
