Question Number 193461 by SAMIRA last updated on 14/Jun/23
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\sqrt{\mathrm{1}+\mathrm{sin}\:\mathrm{x}}−\mathrm{1}}{\mathrm{sin}\:\mathrm{2x}}\right)\:=\:?? \\ $$
Answered by aba last updated on 14/Jun/23
$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}+\mathrm{sinx}−\mathrm{1}}{\mathrm{sin}\left(\mathrm{2x}\right)\left(\sqrt{\mathrm{1}+\mathrm{sinx}}+\mathrm{1}\right)}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{2cosx}\left(\sqrt{\mathrm{1}+\mathrm{sinx}}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{4}}\:\checkmark \\ $$
Answered by MM42 last updated on 14/Jun/23
$${lim}_{{x}\rightarrow\mathrm{0}} \:\left(\frac{{sinx}}{{sin}\mathrm{2}{x}\left(\sqrt{\mathrm{1}+{sinx}}+\mathrm{1}\right)}\right)=\frac{\mathrm{1}}{\mathrm{4}} \\ $$