Question Number 63918 by raj last updated on 11/Jul/19
![lim_(x→π/2) (([1−tan x/2][1−sin x])/([1+tan x/2][π−2x]^3 ))=?](https://www.tinkutara.com/question/Q63918.png)
$$\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\frac{\left[\mathrm{1}−\mathrm{tan}\:{x}/\mathrm{2}\right]\left[\mathrm{1}−\mathrm{sin}\:{x}\right]}{\left[\mathrm{1}+\mathrm{tan}\:{x}/\mathrm{2}\right]\left[\pi−\mathrm{2}{x}\right]^{\mathrm{3}} }=? \\ $$
Answered by Cmr 237 last updated on 20/Aug/19
$${posons}\:{cette}\:{limite}\:{egale}\:{A} \\ $$$${par}\:{le}\:{developement}\:{limite}\:{we}\:{have}: \\ $$$${aux}\:{voisinage}\:{de}\:\frac{\pi}{\mathrm{2}}, \\ $$$${tan}\left(\frac{{x}}{\mathrm{2}}\right)\frown\mathrm{1}+\left({x}−\frac{\pi}{\mathrm{2}}\right)+{o}\left({x}\right) \\ $$$${sinx}\frown\mathrm{1}−\frac{\left({x}−\frac{\pi}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{2}}+{o}\left({x}\right) \\ $$$${A}\left({x}\right)\frown\frac{−\left({x}−\frac{\pi}{\mathrm{2}}\right)\left(\frac{\left({x}−\frac{\pi}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{2}}\right)}{\left(\mathrm{2}+\left({x}−\frac{\pi}{\mathrm{2}}\right)\right)\left(\pi−\mathrm{2}{x}\right)^{\mathrm{3}} }+\mathrm{o}\left(\mathrm{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\frac{\mathrm{1}}{\mathrm{16}}\left(\pi−\mathrm{2}{x}\right)^{\mathrm{3}} }{\left(\mathrm{2}+\left({x}−\frac{\pi}{\mathrm{2}}\right)\right)\left(\pi−\mathrm{2}{x}\right)^{\mathrm{3}} }+{o}\left({x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\frac{\mathrm{1}}{\mathrm{16}}}{\left(\mathrm{2}+\left({x}−\frac{\pi}{\mathrm{2}}\right)\right)}+{o}\left({x}\right) \\ $$$${now}\:{it}'{s}\:{clear}\:{that}; \\ $$$${lim}\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\:}\:{A}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{32}}={A}. \\ $$