Squeeze Theorem

AP Calculus AB· difficulty 2/5

\displaystyle\lim_{x \to 0}\dfrac{\sin(5x)}{x}

A
$\tfrac{1}{5}$
B
$1$
C
$0$
D
$5$
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Explanation

\dfrac{\sin(5x)}{x} = 5 \cdot \dfrac{\sin(5x)}{5x}

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