# Limit of goniometric function without l'Hospital's rule

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

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

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

Set $$t=\frac \pi2 - x,$$ $$\lim_{x\to {\pi\over 2}} \frac {1-\sin x}{(\frac\pi2 -x)^2}=\lim_{t\to {0}} \frac {1-\cos t}{t^2}=\lim_{t\to {0}} \frac {2 \sin ^2(t/2)}{4(t/2)^2}={1\over 2}$$
$$\lim_{x\to \frac{\pi}2} \frac {1-\sin x}{\left(\frac\pi2 -x\right)^2 } =\lim_{x\to \frac{\pi}2} \frac {1-\sin x}{\left(\frac\pi2 -x\right)^2 } \frac {1+\sin x}{1+\sin x} =\lim_{x\to \frac{\pi}2} \frac {1}{1+\sin x}\frac {\sin^2\left(\frac\pi2 -x\right)}{\left(\frac\pi2 -x\right)^2 } =\frac12$$
• @LukášFrajt The second fraction is in the form $y\to 0 \quad \frac{\sin^2 y}{y^2} \to 1$. – user Nov 20 '18 at 23:09
• @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. – user Nov 20 '18 at 23:10