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I'm struggling with the following limit:

$$\lim_{x \to 0} \frac{1-(\cos x)^{\sin x}}{x^2}$$

Don't know where to start with this. Hints/solutions very appreciated.

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  • $\begingroup$ l'Hopital twice. $\endgroup$ – Henricus V. Apr 2 '16 at 16:17
  • $\begingroup$ Thx. It works! Wonder, whether there is another method of finding this limit without using l'Hopital. $\endgroup$ – UCL Apr 2 '16 at 16:18
  • $\begingroup$ I thought of series expansion, but algebra is nasty then... Any help very appreciated. $\endgroup$ – UCL Apr 2 '16 at 16:18
  • $\begingroup$ Series would be a good idea. Start with $\sin(x)\log(cos(x))$ $\endgroup$ – Claude Leibovici Apr 2 '16 at 16:20
  • $\begingroup$ $\frac{1-(\cos x)^{\sin x}}{x^2}=\frac{x}{2}+o(x^3)$ $\endgroup$ – Gabriel Romon Apr 2 '16 at 16:23
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For a series approach, just look at $y(x) = \cos(x)^{\sin(x)}$ and consider the logarithm $\ln y = \sin(x) \ln \cos(x)$. For $x \to 0$ then $\ln y \approx (x + x^3/3) \ln (1 - x^2/2)$ and the expansion for the $\ln$ term is well known.

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We have \begin{align} L &= \lim_{x \to 0}\frac{1 - (\cos x)^{\sin x}}{x^{2}}\notag\\ &= \lim_{x \to 0}\frac{1 - \exp(\sin x\log \cos x)}{x^{2}}\notag\\ &= -\lim_{x \to 0}\frac{\exp(\sin x\log \cos x) - 1}{\sin x\log \cos x}\cdot\frac{\sin x\log \cos x}{x^{2}}\notag\\ &= -\lim_{x \to 0}\frac{\log\cos x}{x}\notag\\ &= -\lim_{x \to 0}\frac{\log\cos^{2} x}{2x}\notag\\ &= -\frac{1}{2}\lim_{x \to 0}\frac{\log(1 - \sin^{2} x)}{\sin^{2}x}\cdot\frac{\sin^{2}x}{x^{2}}\cdot x\notag\\ &= -\frac{1}{2}\cdot(-1)\cdot 1\cdot 0\notag\\ &= 0\notag \end{align} Like most of the limit problems seen on MSE, this one is also evaluated using standard limits without the use of advanced tools like L'Hospital's Rule or Taylor's series.

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