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Establish the given limit formula

by RoRi

Prove the formula for the limit:

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

Using,

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

Proof. Since \tan (2x) = \frac{\sin (2x)}{\cos (2x)} we have,

    \begin{align*} \lim_{x \to 0} \frac{\tan (2x)}{\sin x} &= \lim_{x \to 0} \frac{\sin (2x)}{\cos(2x) \sin x} \\ &= \lim_{x \to 0} \frac{2 \sin x \cos x}{\cos (2x) \sin x} &(\sin (2x) = 2 \sin x \cos x) \\ &= \lim_{x \to 0} \frac{ 2 \cos x}{\cos (2x)} \\ &= \frac{2}{1} = 2. \end{align*}

Where the final line follows since \frac{ 2 \cos x}{\cos (2x)} is a quotient of continuous functions, and \cos (2x) \neq 0 at x = 0, so the limit is just the value at x = 0. \qquad \blacksquare

One comment

  1. Anonymous says:

    The special limit (i.e., (sin x)/x approaches 1 as x approaches 0) is not used in this solution. Perhaps Apostol did not mean the header above Exercises 15-20 all too literally. You might, or might not, find the special limit useful.

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