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I try with taylor expansion to prove that $$\lim_{\theta\to0}\frac{\cos\left(\frac\pi2\cos\theta\right)}{\sin\theta}=0$$ but I have always a singularity when $\sin\theta\approx\theta$

Any help is welcome

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    $\begingroup$ Does $a$ happen to be something like $\frac \pi2$? Otherwise, the numerator tends to $\cos a\ne 0$ and the denominator tends to $0$ $\endgroup$ – Hagen von Eitzen Oct 26 '18 at 10:55
  • $\begingroup$ -@Hagen von Eitzen . yes I forgot to say that $a$ is equal to $\pi/2$ $\endgroup$ – youpilat13 Oct 26 '18 at 10:56
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Taylor expansion gives you

$$\cos\left(\frac{\pi}{2}\cos\theta\right)\approx \cos\left(\frac{\pi}{2}(1-\theta^2/2)\right)\approx \cos\left(\frac{\pi}{2}-\frac{\pi}{4}\theta^2\right)=\sin\left(\frac{\pi}{4}\theta^2\right)\approx \frac{\pi}{4}\theta^2 $$ while the denominator is just $\sin\theta\approx \theta$. You see that the numerator has a higher leading order and thus overpowers the denominator, resulting in zero limit.

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  • $\begingroup$ thanks for all ! $\endgroup$ – youpilat13 Oct 26 '18 at 14:28
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The numerator is $\sin(\pi\sin^2\frac{\theta}{2})\approx\frac{\pi\theta^2}{4}$, while the denominator is approximately $\theta$, giving a ratio $\approx\frac{\pi\theta}{4}\to 0$. This approach requires $\lim_{\theta\to 0}\frac{\sin\theta}{\theta}=1$, but not more advanced techniques such as Taylor series or L'Hôpital's rule. More formally, we have $$\lim_{\theta\to 0}\frac{\sin(\pi\sin^2\frac{\theta}{2})}{\pi\sin^2\frac{\theta}{2}}\frac{\pi\sin^2\frac{\theta}{2}}{(\frac{\theta}{2})^2}\frac{(\frac{\theta}{2})^2}{\theta}\frac{\theta}{\sin\theta}=1\cdot\pi\cdot 0\cdot1=0.$$

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HINT: Apply L'Hopital's Rule: $$\lim _{θ\to 0}\left(\frac{\frac{\pi }{2}\sin \left(\frac{\pi }{2}\cos \left(θ\right)\right)\sin \left(θ\right)}{\cos \left(θ\right)}\right)$$

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