Question Number 84460 by jagoll last updated on 13/Mar/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(\mathrm{2}+\mathrm{x}\right)−\mathrm{sin}\:\left(\mathrm{2}−\mathrm{x}\right)}{\mathrm{x}} \\ $$
Commented by jagoll last updated on 13/Mar/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left(\mathrm{2}+\mathrm{x}\right)+\mathrm{cos}\:\left(\mathrm{2}−\mathrm{x}\right)}{\mathrm{1}} \\ $$$$=\:\mathrm{2cos}\:\mathrm{2} \\ $$
Commented by mathmax by abdo last updated on 13/Mar/20
$${let}\:{f}\left({x}\right)=\frac{{sin}\left(\mathrm{2}+{x}\right)−{sin}\left(\mathrm{2}−{x}\right)}{{x}}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\frac{{sin}\mathrm{2}\:{cosx}\:+{cos}\mathrm{2}{sinx}−\left({sin}\mathrm{2}\:{cosx}\:−{cos}\mathrm{2}\:{sinx}\right)}{{x}} \\ $$$$=\frac{\mathrm{2}{cos}\mathrm{2}\:{sinx}}{{x}}\:\sim\:\mathrm{2}{cos}\mathrm{2}\:\:\:\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} {f}\left({x}\right)=\mathrm{2}{cos}\mathrm{2} \\ $$