Question Number 114259 by bobhans last updated on 18/Sep/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\:\mathrm{arc}\:\mathrm{sin}\:\left({x}^{\mathrm{2}} \right)}{{x}\:\mathrm{cos}\:{x}−\mathrm{sin}\:{x}}\:? \\ $$
Answered by bemath last updated on 18/Sep/20
$${by}\:{L}'{Hopital} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{arc}\:\mathrm{sin}\:\left({x}^{\mathrm{2}} \right)+\frac{\mathrm{2}{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}}{\mathrm{cos}\:{x}−{x}\:\mathrm{sin}\:{x}\:−\mathrm{cos}\:{x}}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{arc}\:\mathrm{sin}\:\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }\:+\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}}{−\frac{\mathrm{sin}\:{x}}{{x}}}\:= \\ $$$${now}\:{let}\:{L}_{\mathrm{1}} \:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{arc}\:\mathrm{sin}\:\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} } \\ $$$${L}_{\mathrm{1}} =\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}}{\mathrm{2}{x}}\:=\:\mathrm{1} \\ $$$${so}\:{L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{arc}\:\mathrm{sin}\:\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }\:+\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}}{−\:\frac{\mathrm{sin}\:{x}}{{x}}} \\ $$$${L}\:=\:\frac{\mathrm{1}+\mathrm{2}}{−\mathrm{1}}\:=\:−\mathrm{3} \\ $$
Answered by Olaf last updated on 18/Sep/20
$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}×{x}^{\mathrm{2}} }{{x}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)−\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\right)} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}^{\mathrm{3}} }{−\frac{{x}^{\mathrm{3}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{6}}}\:=\:\frac{\mathrm{1}}{−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{6}}}\:=\:−\mathrm{3} \\ $$$$ \\ $$
Commented by bemath last updated on 18/Sep/20
$${santuyy} \\ $$
Commented by malwaan last updated on 18/Sep/20
$${verysantuyygood} \\ $$
Commented by abdullahquwatan last updated on 18/Sep/20
$${mantep} \\ $$
Answered by mathmax by abdo last updated on 19/Sep/20
$$\mathrm{we}\:\mathrm{have}\:\mathrm{xarcsin}\left(\mathrm{x}^{\mathrm{2}} \right)\sim\mathrm{x}^{\mathrm{3}} \\ $$$$\mathrm{cosx}\:\sim\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\mathrm{xcosx}\:\sim\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{2}}\:\:\mathrm{and}\:\mathrm{sinx}\:\sim\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\sim\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{2}}−\mathrm{x}\:+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}}\:=\frac{\mathrm{x}^{\mathrm{3}} }{−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}^{\mathrm{3}} }\:=−\mathrm{3}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{x}\right)\:=−\mathrm{3} \\ $$