L'Hôpital's Rule

AP Calculus AB· difficulty 2/5

\displaystyle\lim_{x\to 0}\dfrac{\sin(2x)}{\tan(3x)}

A
$0$
B
$\tfrac{2}{3}$
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C
$\tfrac{3}{2}$
D
$1$

Explanation

2\cos(2x)/(3\sec^2(3x)) \to 2/3

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