Question Number 157191 by naka3546 last updated on 20/Oct/21
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{sin}\:{x}\:−\:{x}\:+\:\frac{\mathrm{1}}{\mathrm{6}}\:{x}^{\mathrm{3}} }{{x}^{\mathrm{5}} }\:\:\:=\:\:? \\ $$$$\left(\:{Without}\:\:{L}'{Hospital}\:,\:{Taylor}\:\:{or}\:\:{Maclaurin}\:\:{Series}\:\right)\:. \\ $$
Answered by puissant last updated on 20/Oct/21
$${We}\:{have}\:{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}\:\leqslant\:{sinx}\:\leqslant\:{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!} \\ $$$$\Rightarrow\:\mathrm{0}\:\leqslant\:{sinx}−{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\:\leqslant\:\frac{{x}^{\mathrm{5}} }{\mathrm{120}} \\ $$$$\Rightarrow\:\mathrm{0}\:\leqslant\:\frac{{sinx}−{x}+\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} }{{x}^{\mathrm{5}} }\:\leqslant\:\frac{{x}^{\mathrm{5}} }{\mathrm{120}{x}^{\mathrm{5}} }=\frac{\mathrm{1}}{\mathrm{120}} \\ $$$$\Rightarrow\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{sinx}−{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{6}}}{{x}^{\mathrm{5}} }\:=\:\frac{\mathrm{1}}{\mathrm{120}}.. \\ $$
Commented by naka3546 last updated on 20/Oct/21
$${Thank}\:\:{you}\:. \\ $$
Commented by Le_Professeur last updated on 20/Oct/21
$${without}\:{taylor}\:{or}\:{mac}\:{laurin}\:{series}\left({without}\:{c}\mathrm{7}\:{sans}\:{utiliser}\right) \\ $$$$ \\ $$