Question Number 114637 by bobhans last updated on 20/Sep/20 | ||
$$\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\frac{\pi}{\mathrm{2}}−\mathrm{cos}\:^{−\mathrm{1}} \left(\mathrm{2}{x}−\pi\right)}{\mathrm{1}−\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}}{\pi}\right)}\:? \\ $$ | ||
Answered by bemath last updated on 20/Sep/20 | ||
$${setting}\:{x}=\frac{\pi}{\mathrm{2}}+{p}\rightarrow\mathrm{2}{x}=\pi+\mathrm{2}{p} \\ $$$$\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\pi}{\mathrm{2}}−\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{2}{p}\right)}{\mathrm{1}−\mathrm{sin}^{−\mathrm{1}} \left(\frac{\pi+\mathrm{2}{p}}{\pi}\right)}= \\ $$$$\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left[\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−\mathrm{4}{p}^{\mathrm{2}} }}\right]}{−\left[\frac{\frac{\mathrm{2}}{\pi}}{\:\sqrt{\mathrm{1}−\left(\frac{\pi+\mathrm{2}{p}}{\pi}\right)^{\mathrm{2}} }}\right]}\:= \\ $$$$\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−\mathrm{4}{p}^{\mathrm{2}} }}\:×\:\left(−\frac{\mathrm{2}}{\pi\sqrt{\mathrm{1}−\left(\frac{\pi+\mathrm{2}{p}}{\pi}\right)^{\mathrm{2}} }}\right)\:=\:\infty \\ $$$$ \\ $$ | ||
Commented by bobhans last updated on 20/Sep/20 | ||
$${gave}\:{kudos} \\ $$ | ||