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Soal 11

Nomor 11

Nilai dari $\lim_{x \to \frac{\pi}{2}} \frac{\sec 2x + 2}{\tan 2x}$ adalah....

$-2$

$-1$

$-\frac{1}{2}$

$0$

$1$

Pembahasan

Misalkan $y = x - \frac{\pi}{2}$ maka $x = y + \frac{\pi}{2}$ . Selesaikan limit dengan permisalan di atas.

= \lim_{y \to 0} \frac{\sec 2\left(y + \frac{\pi}{2}\right) + 2}{\tan 2\left(y + \frac{\pi}{2}\right)}

= \lim_{y \to 0} \frac{-\sec 2y + 2}{\tan 2y}

= \lim_{y \to 0} \frac{-\frac{1}{\cos 2y} + 2}{\frac{\sin 2y}{\cos 2y}}

= \lim_{y \to 0} \frac{\frac{-1 + 2 \cos 2y}{\cos 2y}}{\frac{\sin 2y}{\cos 2y}}

= \lim_{y \to 0} \frac{-1 + 2 \cos 2y}{\sin 2y}

Gunakan konsep L'Hospital

= \lim_{y \to 0} \frac{-4 \sin 2y}{2 \cos 2y}

= \lim_{y \to 0} -2 \frac{\sin 2y}{\cos 2y}

= \lim_{y \to 0} -2 \tan 2y

= -2 \tan(2 \cdot 0)

= -2 \tan 0

= -2 \cdot 0

= 0

Komentar

Soal 11Set 3

Latihan

Nomor 11

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