Question Number 126761 by bemath last updated on 24/Dec/20
Answered by liberty last updated on 24/Dec/20
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{3}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{x}−\frac{\mathrm{1}}{\mathrm{8}}{x}^{\mathrm{2}} +{O}\left({x}^{\mathrm{3}} \right)\right)−\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}+{O}\left({x}^{\mathrm{5}} \right)\right)^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{3}} \left({x}+\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} +{O}\left({x}^{\mathrm{5}} \right)\right)}{\left(\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{3}} +{O}\left({x}^{\mathrm{6}} \right)\right)−\mathrm{1}\right)\left({x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{4}} +{O}\left({x}^{\mathrm{6}} \right)\right)}= \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}^{\mathrm{5}} \left(\frac{\mathrm{3}}{\mathrm{8}}+{O}\left({x}\right)\right)}{{x}^{\mathrm{5}} \left(\frac{\mathrm{2}}{\mathrm{3}}+{O}\left({x}\right)\right)}\:=\:\frac{\mathrm{3}}{\mathrm{8}}×\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{9}}{\mathrm{16}} \\ $$