Question Number 137820 by bramlexs22 last updated on 07/Apr/21
$$\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}−\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right)\left(\mathrm{1}−\mathrm{sin}\:{x}\right)}{\left(\mathrm{1}+\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right)\left(\pi−\mathrm{2}{x}\right)^{\mathrm{3}} }\:? \\ $$
Answered by SLVR last updated on 07/Apr/21
Commented by SLVR last updated on 07/Apr/21
$${is}\:{it}\:{okay} \\ $$$$ \\ $$
Commented by bramlexs22 last updated on 07/Apr/21
$${answer}\:\frac{\mathrm{1}}{\mathrm{32}} \\ $$
Commented by SLVR last updated on 07/Apr/21
$${i}\:{am}\:{sorry}..{caliculation}…{error} \\ $$$${see}\:{this}\:..{but}\:\mathrm{1}/\mathrm{32}\:{i}\:{hope} \\ $$
Commented by SLVR last updated on 07/Apr/21
Answered by liberty last updated on 07/Apr/21
$$\:\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}\right)\left(\mathrm{1}−\mathrm{sin}\:{x}\right)}{\left(\pi−\mathrm{2}{x}\right)^{\mathrm{3}} } \\ $$$${let}\:{x}=\frac{\pi}{\mathrm{2}}+{z}\:\&\:{z}\rightarrow\mathrm{0} \\ $$$$\begin{cases}{\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}=−\frac{{z}}{\mathrm{2}}}\\{\mathrm{1}−\mathrm{sin}\:{x}\:=\:\mathrm{1}−\mathrm{cos}\:{z}}\\{\pi−\mathrm{2}{x}\:=\:−\mathrm{8}{z}^{\mathrm{3}} }\end{cases} \\ $$$${so}\:\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{tan}\:\left(\frac{{z}}{\mathrm{2}}\right)\left(\mathrm{1}−\mathrm{cos}\:{z}\right)}{−\mathrm{8}{z}^{\mathrm{3}} }= \\ $$$$\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\frac{{z}}{\mathrm{2}}\right)\left(\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{{z}}{\mathrm{2}}\right)\right.}{\mathrm{8}{z}^{\mathrm{3}} }\:=\:\frac{\mathrm{1}}{\mathrm{32}} \\ $$