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Find the limit as x goes to 0 of (cosh x – cos x) / x2

Evaluate the limit.

    \[ \lim_{x \to 0} \frac{\cosh x - \cos x}{x^2}. \]


We use the definition of \cosh x in terms of the exponential:

    \[ \cosh x = \frac{e^x + e^{-x}}{2}, \]

and the expansions (page 287 of Apostol) of e^x and \cos x as x \to 0:

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

Putting these together we evaluate the limit:

    \begin{align*}  \lim_{x \to 0} \frac{\cosh x - \cos x}{x^2} &= \lim_{x \to 0} \frac{e^x + e^{-x} - 2\cos x}{2x^2} \\[9pt]  &= \lim_{x \to 0} \frac{1 + x + \frac{x^2}{2} + \frac{x^3}{6} + o(x^3) + 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + o(x^3) - 2 \cos x}{2x^2} \\[9pt]  &= \lim_{x \to 0} \frac{2 + x^2 + o(x^3) - 2\left(1 - \frac{x^2}{2} + o(x^3) \right)}{2x^2} \\[9pt]  &= \lim_{x \to 0} \frac{2x^2 + o(x^3)}{2x^2} \\[9pt]  &= \lim_{x \to 0} \left( 1 + \frac{o(x^3)}{x^2} \right) \\[9pt]  &= 1. \end{align*}

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