Question Number 195148 by horsebrand11 last updated on 25/Jul/23
$$\:\:\begin{array}{|c|}{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}+\mathrm{x}\:\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}=?}\\\hline\end{array} \\ $$
Answered by Erico last updated on 25/Jul/23
$$\frac{\mathrm{1}+\mathrm{xsinx}−\mathrm{cosx}}{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}=\frac{\mathrm{1}−\mathrm{cosx}}{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}+\frac{\mathrm{x}}{\mathrm{sinx}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}−\mathrm{cosx}}{\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \mathrm{x}\right)}+\frac{\mathrm{x}}{\mathrm{sinx}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cosx}}+\frac{\mathrm{x}}{\mathrm{sinx}}\:\:\:\:\underset{\mathrm{x}\rightarrow\mathrm{0}} {\rightarrow}\:\:\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Answered by dimentri last updated on 25/Jul/23
$$\:\:\:\:\underbrace{\Subset} \\ $$
Answered by SajaRashki last updated on 25/Jul/23
$$\overset{{hop}} {=}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}{sinx}+{xcosx}}{{sin}\mathrm{2}{x}}\:\:\overset{{hop}} {=}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{3}{cosx}−{xsinx}}{\mathrm{2}{cosx}}= \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\:\blacksquare \\ $$