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Establish the given limit formula

by RoRi

Prove the formula for the limit:

    \[ \lim_{x \to 0} \frac{\sin (2x)}{x} = 2. \]

Using,

    \[ \lim_{x \to 0} \frac{\sin x}{x} = 1. \]

Proof. First, we recall the trig identity:

    \[ \sin (2x) = 2 \sin x \cos x. \]

So,

    \[ \frac{\sin (2x)}{x} = \frac{2 \sin x \cos x}{x} = 2 \cos x \left( \frac{\sin x}{x} \right). \]

Each limit in the product exists as x \to 0, so the limit of the product is the product of the limits (by Theorem 3.1 (iii)):

    \begin{align*} \lim_{x \to 0} \frac{\sin (2x)}{x} &= \lim_{x \to 0} \left( 2 \cos x \frac{\sin x}{x} \right) \\ &= \left( \lim_{x \to 0} 2 \right) \left( \lim_{x \to 0} \cos x \right) \left( \lim_{x \to 0} \frac{\sin x}{x} \right) \\ &= 2 \cdot 1 \cdot 1 \\ &= 2. \qquad \blacksquare \end{align*}

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