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I'm trying figure this out without l'Hospital's rule. But I don't know how should I start. Any hint, please?

$$\lim_{x\to \frac{\pi}2} \frac {1-\sin x}{\left(\frac\pi2 -x\right)^2 }$$

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  • $\begingroup$ Your equation makes no sense. $\sin$ needs an argument, at the very least. $\endgroup$ – David G. Stork Nov 20 '18 at 22:31
  • $\begingroup$ Why isn't this trivially $0$? You have a constant in the denominator (which can be removed) and the numerator is clearly $0$. $\endgroup$ – David G. Stork Nov 20 '18 at 22:33
  • $\begingroup$ answer in the book is 1/2. But I cannot solve it. Also in wolfram alpha.. $\endgroup$ – Lukáš Frajt Nov 20 '18 at 22:34
  • $\begingroup$ Shouldn't be the $1$ in the denominator replaced by $x$? $\endgroup$ – user376343 Nov 20 '18 at 22:47
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Set $t=\frac \pi2 - x,$ $$\lim_{x\to {\pi\over 2}} \frac {1-\sin x}{(\frac\pi2 -x)^2}=\lim_{t\to {0}} \frac {1-\cos t}{t^2}=\lim_{t\to {0}} \frac {2 \sin ^2(t/2)}{4(t/2)^2}={1\over 2}$$

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We have that

$$\lim_{x\to \frac{\pi}2} \frac {1-\sin x}{\left(\frac\pi2 -x\right)^2 } =\lim_{x\to \frac{\pi}2} \frac {1-\sin x}{\left(\frac\pi2 -x\right)^2 } \frac {1+\sin x}{1+\sin x} =\lim_{x\to \frac{\pi}2} \frac {1}{1+\sin x}\frac {\sin^2\left(\frac\pi2 -x\right)}{\left(\frac\pi2 -x\right)^2 } =\frac12$$

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  • $\begingroup$ Is that right? we still have 0/0 in the second fraction $\endgroup$ – Lukáš Frajt Nov 20 '18 at 23:07
  • $\begingroup$ @LukášFrajt The second fraction is in the form $y\to 0 \quad \frac{\sin^2 y}{y^2} \to 1$. $\endgroup$ – user Nov 20 '18 at 23:09
  • $\begingroup$ @LukášFrajt The simpler way is note that $$\lim_{x\to \frac{\pi}2} \frac {1-\sin x}{\left(\frac\pi2 -x\right)^2 }=\lim_{x\to \frac{\pi}2} \frac {1-\cos \left(\frac\pi2 -x\right)}{\left(\frac\pi2 -x\right)^2 }=\frac12$$ by standard limits. $\endgroup$ – user Nov 20 '18 at 23:10
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    $\begingroup$ Got it, thank you! $\endgroup$ – Lukáš Frajt Nov 20 '18 at 23:11

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