Question Number 109722 by bemath last updated on 25/Aug/20
$$\:\:\:\:\:\multimap\frac{\flat\epsilon}{{math}}\multimap \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:{x}}{\:\sqrt{\mathrm{1}−\mathrm{cos}\:{x}}}\:=\:?\: \\ $$
Commented by bemath last updated on 25/Aug/20
Answered by mathmax by abdo last updated on 25/Aug/20
$$\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{sinx}}{\:\sqrt{\mathrm{1}−\mathrm{cosx}}}\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{1}−\mathrm{cosx}\:\sim\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\sqrt{\mathrm{1}−\mathrm{cosx}}\sim\frac{\mathrm{x}}{\:\sqrt{\mathrm{2}}} \\ $$$$\mathrm{akso}\:\mathrm{sinx}\:\sim\mathrm{x}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:\sim\frac{\mathrm{x}}{\frac{\mathrm{x}}{\:\sqrt{\mathrm{2}}}}\:=\sqrt{\mathrm{2}}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{2}} \\ $$
Answered by mathmax by abdo last updated on 25/Aug/20
$$\mathrm{here}\:\mathrm{x}\rightarrow\mathrm{0}^{+} \\ $$