Question Number 187645 by Mingma last updated on 19/Feb/23
Answered by Ar Brandon last updated on 19/Feb/23
$$\mathscr{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}^{\mathrm{4}} \left(\pi\mathrm{cos}{x}\right)}{\mathrm{1}−\mathrm{cos}\left(\mathrm{1}−\mathrm{cos}\left(\mathrm{1}−\mathrm{cos}{x}\right)\right)} \\ $$$$\:\:\:\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}^{\mathrm{4}} \left(\pi−\frac{\pi{x}^{\mathrm{2}} }{\mathrm{2}}\right)}{\frac{{x}^{\mathrm{8}} }{\mathrm{128}}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}^{\mathrm{4}} \left(\frac{\pi{x}^{\mathrm{2}} }{\mathrm{2}}\right)}{\frac{{x}^{\mathrm{8}} }{\mathrm{128}}} \\ $$$$\:\:\:\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{sin}\left(\frac{\pi{x}^{\mathrm{2}} }{\mathrm{2}}\right)}{\frac{\pi{x}^{\mathrm{2}} }{\mathrm{2}}}\right)^{\mathrm{4}} ×\mathrm{8}\pi^{\mathrm{4}} =\mathrm{8}\pi^{\mathrm{4}} \\ $$