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Find the limit as x goes to 0 of (cos (sin x) – cos x) / x4

Evaluate the limit.

    \[ \lim_{x \to 0} \frac{\cos (\sin x) - \sin x}{x^4}. \]


We know (page 287 of Apostol) that the expansions for \sin x and \cos x as x \to 0 are given by

    \begin{align*}   \sin x &= x - \frac{x^3}{6} + o(x^4) \\  \cos x &= 1 - \frac{x^2}{2} + \frac{x^4}{24} + o(x^5).  \end{align*}

Therefore, we have the following expansion for \cos (\sin x) as x \to 0,

    \begin{align*}  \cos (\sin x) &= 1 - \frac{(\sin x)^2}{2} + \frac{(\sin x)^4}{24} + o((\sin x)^5) \\[9pt]  &= 1 - \frac{1}{2} \left( x - \frac{x^3}{6} + o(x^4) \right)^2 + \frac{1}{24} \left( x - \frac{x^3}{6} + o(x^4) \right)^4 + o(x^5) \\[9pt]  &= 1 - \frac{1}{2} \left( x^2 - \frac{x^4}{3} + o(x^4) \right) + \frac{1}{24} \left(x^4 + o(x^4)\right) + o(x^5) \\[9pt]  &= 1 - \frac{1}{2}x^2 + \frac{5}{24}x^4 + o(x^4). \end{align*}

So, now we can take the limit,

    \begin{align*}  \lim_{x \to 0} \frac{\cos (\sin x) - \sin x}{x^4} &= \lim_{x \to 0} \frac{1 - \frac{1}{2}x^2 + \frac{5}{24}x^4 + o(x^4) - 1 + \frac{1}{2}x^2 - \frac{1}{24}x^4 + o(x^5)}{x^4} \\[9pt]  &= \lim_{x \to 0} \frac{\frac{1}{6} x^4 + o(x^4)}{x^4} \\[9pt]  &= \lim_{x \to 0} \left( \frac{1}{6} + \frac{o(x^4)}{x^4} \right) \\[9pt]  &= \frac{1}{6}. \end{align*}

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