Question Number 165054 by mathlove last updated on 25/Jan/22
Answered by alephzero last updated on 25/Jan/22
$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\left(\sqrt{\mathrm{sin}\:\frac{\pi}{{x}}}\::\:\mathrm{cos}\:\frac{\pi}{{x}+\mathrm{2}}\right)^{\frac{\mathrm{1}}{{x}}} \:= \\ $$$$=\:\sqrt{\sqrt{\mathrm{sin}\:\frac{\pi}{\mathrm{2}}}\::\:\mathrm{cos}\:\frac{\pi}{\mathrm{4}}}\:= \\ $$$$=\:\sqrt{\mathrm{1}\::\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\:=\:\sqrt{\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}}}\:=\:\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\sqrt{\mathrm{2}}}}\:=\:\frac{\sqrt{\mathrm{2}}}{\:\sqrt[{\mathrm{4}}]{\mathrm{2}}}\:=\: \\ $$$$=\:\frac{\sqrt{\mathrm{2}}\sqrt[{\mathrm{4}}]{\mathrm{2}^{\mathrm{3}} }}{\mathrm{2}}\:=\:\frac{\sqrt[{\mathrm{4}}]{\mathrm{2}^{\mathrm{5}} }}{\mathrm{2}}\:=\:\frac{\mathrm{2}\sqrt[{\mathrm{4}}]{\mathrm{2}}}{\mathrm{2}}\:=\:\sqrt[{\mathrm{4}}]{\mathrm{2}} \\ $$