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Question Number 126761 by bemath last updated on 24/Dec/20

Answered by liberty last updated on 24/Dec/20

lim_(x→0) ((x^3 (1+(1/2)x−(1/8)x^2 +O(x^3 ))−(x−(x^3 /6)+O(x^5 ))^3 −(1/2)x^3 (x+(1/3)x^3 +O(x^5 )))/(((1+(2/3)x^3 +O(x^6 ))−1)(x^2 −(1/2)x^4 +O(x^6 ))))= lim_(x→0) ((x^5 ((3/8)+O(x)))/(x^5 ((2/3)+O(x)))) = (3/8)×(3/2)=(9/(16))

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{3}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{x}−\frac{\mathrm{1}}{\mathrm{8}}{x}^{\mathrm{2}} +{O}\left({x}^{\mathrm{3}} \right)\right)−\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}+{O}\left({x}^{\mathrm{5}} \right)\right)^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{3}} \left({x}+\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} +{O}\left({x}^{\mathrm{5}} \right)\right)}{\left(\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{3}} +{O}\left({x}^{\mathrm{6}} \right)\right)−\mathrm{1}\right)\left({x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{4}} +{O}\left({x}^{\mathrm{6}} \right)\right)}= \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}^{\mathrm{5}} \left(\frac{\mathrm{3}}{\mathrm{8}}+{O}\left({x}\right)\right)}{{x}^{\mathrm{5}} \left(\frac{\mathrm{2}}{\mathrm{3}}+{O}\left({x}\right)\right)}\:=\:\frac{\mathrm{3}}{\mathrm{8}}×\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{9}}{\mathrm{16}} \\ $$

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