Question Number 124915 by liberty last updated on 06/Dec/20
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2tan}\:{x}−\mathrm{5sin}\:{x}+\mathrm{3}{x}−{x}^{\mathrm{3}} }{\mathrm{1}−\sqrt[{\mathrm{5}}]{\mathrm{1}−\mathrm{2}{x}^{\mathrm{3}} }}\:=?\: \\ $$
Answered by bemath last updated on 07/Dec/20
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\left({x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right)−\mathrm{5}\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\right)+\mathrm{3}{x}−{x}^{\mathrm{3}} }{\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{2}{x}^{\mathrm{3}} }{\mathrm{5}}\right)}\:= \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{2}{x}^{\mathrm{3}} }{\mathrm{3}}\:+\:\frac{\mathrm{5}{x}^{\mathrm{3}} }{\mathrm{6}}−{x}^{\mathrm{3}} }{\frac{\mathrm{2}{x}^{\mathrm{3}} }{\mathrm{5}}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{3}} }{\frac{\mathrm{2}{x}^{\mathrm{3}} }{\mathrm{5}}}\:=\:\frac{\mathrm{5}}{\mathrm{4}} \\ $$