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Find the limit as x goes to 0 of the given expression

by RoRi

Evaluate the limit.

    \[ \lim_{x \to 0} \frac{\log(1+x) - x}{1 - \cos x}. \]

We use the expansions on page 287 for \log(1+x) and \cos x,

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

Then we compute the limit,

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

One comment

  1. tom says:

    My intuition was this one diverges. I was thinking cosx near 0 acts like x^2 (from Pythagorean) whereas log @1 acts like x. Indeed, log(1+x)/1-cosx diverges, and so does x/1-cosx . Very curious, this one…..

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