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Evaluate the limit as x goes to 0 of a given function

by RoRi

Evaluate the limit.

    \[ \lim_{x \to 0} \frac{3 \tan (4x) - 12 \tan x}{3 \sin (4x) - 12 \sin x}. \]

We use the expansions for \tan x (given in Example 1 on page 288 of Apostol) and \sin x (on page 287 of Apostol):

    \begin{align*} \tan x &= x + \frac{1}{3}x^3 + o(x^3) \\ \tan (4x) &= 4x + \frac{64}{3}x^3 + o(x^3) \\ \sin x &= x - \frac{1}{6}x^3 + o(x^3) \\ \sin (4x) &= 4x - \frac{32}{3}x^3 + o(x^3). \end{align*}

Therefore, we have

    \begin{align*} \lim_{x \to 0} \frac{3 \tan (4x) - 12 \tan x}{3 \sin (4x) - 12 \sin x} &= \lim_{x \to 0} \frac{3 \left( 4x + \frac{64}{3} x^3 + o(x^3) \right) - 12 \left( x + \frac{1}{3} x^3 + o(x^3) \right)}{ 3 \left( 4x - \frac{32}{3} x^3 + o(x^3) \right) - 12 \left( x - \frac{1}{6}x^3 + o(x^3) \right)} \\[10pt] &= \lim_{x \to 0} \frac{12x + 64x^3 + o(x^3) - 12x - 4x^3 + o(x^3)}{12x - 32x^3 + o(x^3) - 12x + 2x^3 + o(x^3)} \\[10pt] &= \lim_{x \to 0} \frac{60x^3 + o(x^3)}{-30x^3 + o(x^3)} \\[10pt] &= \lim_{x \to 0} \frac{60 + \frac{o(x^3)}{x^3}}{-30 + \frac{o(x^3)}{x^3}} \\[10pt] &= -2. \end{align*}

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