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Establish the given limit formula

by RoRi

Prove the formula for the limit:

    \[ \lim_{x \to 0} \frac{\sin (5x) - \sin (3x)}{x} = 2. \]

Using,

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

Proof. We know from the previous exercise (Section 3.6, #17),

    \[ \lim_{x \to 0} \frac{\sin (5x)}{x} = 5. \]

Then,

    \begin{align*} \lim_{x \to 0} \frac{\sin (3x)}{x} &= \lim_{x \to 0} \frac{\sin (2x+x)}{x} \\ &= \lim_{x \to 0} \frac{\sin (2x) \cos x + \sin x \cos (2x)}{x} \\ &= \lim_{x \to 0} \left( \frac{\sin (2x)}{x} \cos x + \frac{\sin x}{x} \cos (2x) \right). \end{align*}

But we know from another earlier exercise (Section 3.6, #15) that

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

So, we have,

    \[ \lim_{x \to 0} \frac{\sin (3x)}{x} = 2 + 1 = 3. \]

Hence, by Apostol’s Theorem 3.1 (ii), the limit of the difference is the difference of the limits:

    \begin{align*} \lim_{x \to 0} \frac{\sin (5x) - \sin (3x)}{x} &= \lim_{x \to 0} \left( \frac{\sin (5x)}{x} - \frac{\sin (3x)}{x} \right) \\ &= \lim_{x \to 0} \frac{\sin (5x)}{x} - \lim_{x \to 0} \frac{\sin (3x)}{x} \\ &= 5 - 3 \\ &= 2. \qquad \blacksquare \end{align*}

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