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Question Number 157191 by naka3546 last updated on 20/Oct/21

lim_(x→0)   ((sin x − x + (1/6) x^3 )/x^5 )   =  ?  ( Without  L′Hospital , Taylor  or  Maclaurin  Series ) .

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{sin}\:{x}\:−\:{x}\:+\:\frac{\mathrm{1}}{\mathrm{6}}\:{x}^{\mathrm{3}} }{{x}^{\mathrm{5}} }\:\:\:=\:\:? \\ $$$$\left(\:{Without}\:\:{L}'{Hospital}\:,\:{Taylor}\:\:{or}\:\:{Maclaurin}\:\:{Series}\:\right)\:. \\ $$

Answered by puissant last updated on 20/Oct/21

We have x−(x^3 /(3!)) ≤ sinx ≤ x−(x^3 /(3!))+(x^5 /(5!))  ⇒ 0 ≤ sinx−x+(x^3 /6) ≤ (x^5 /(120))  ⇒ 0 ≤ ((sinx−x+(1/6)x^3 )/x^5 ) ≤ (x^5 /(120x^5 ))=(1/(120))  ⇒ lim_(x→0) ((sinx−x+(x^3 /6))/x^5 ) = (1/(120))..

$${We}\:{have}\:{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}\:\leqslant\:{sinx}\:\leqslant\:{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!} \\ $$$$\Rightarrow\:\mathrm{0}\:\leqslant\:{sinx}−{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\:\leqslant\:\frac{{x}^{\mathrm{5}} }{\mathrm{120}} \\ $$$$\Rightarrow\:\mathrm{0}\:\leqslant\:\frac{{sinx}−{x}+\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} }{{x}^{\mathrm{5}} }\:\leqslant\:\frac{{x}^{\mathrm{5}} }{\mathrm{120}{x}^{\mathrm{5}} }=\frac{\mathrm{1}}{\mathrm{120}} \\ $$$$\Rightarrow\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{sinx}−{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{6}}}{{x}^{\mathrm{5}} }\:=\:\frac{\mathrm{1}}{\mathrm{120}}.. \\ $$

Commented by naka3546 last updated on 20/Oct/21

Thank  you .

$${Thank}\:\:{you}\:. \\ $$

Commented by Le_Professeur last updated on 20/Oct/21

without taylor or mac laurin series(without c7 sans utiliser)

$${without}\:{taylor}\:{or}\:{mac}\:{laurin}\:{series}\left({without}\:{c}\mathrm{7}\:{sans}\:{utiliser}\right) \\ $$$$ \\ $$

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