Question Number 63473 by Rio Michael last updated on 04/Jul/19
$${question} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{sin}\left({x}+{A}\right)−{sin}\left({A}−{x}\right)}{\mathrm{2}{x}} \\ $$
Commented by mathmax by abdo last updated on 04/Jul/19
$${let}\:{A}\left({x}\right)=\frac{{sin}\left({x}+{A}\right)−{sin}\left({A}−{x}\right)}{\mathrm{2}{x}}\:\Rightarrow \\ $$$${A}\left({x}\right)=\frac{{sinx}\:{cosA}\:+{cosx}\:{sinA}\:\:+{sinx}\:{cosA}\:−{cosx}\:{sinA}}{\mathrm{2}{x}} \\ $$$$=\frac{{sinx}}{{x}}\:{cosA}\:\:{due}\:{to}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{{sinx}}{{x}}\:=\mathrm{1}\:\:{we}\:{get}\:{lim}_{{x}\rightarrow\mathrm{0}} \:{A}\left({x}\right)\:={cosA}. \\ $$$${let}\:{use}\:{hospital}\:{theorem} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:{A}\left({x}\right)\:={lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\frac{{cos}\left({x}+{A}\right)\:+{cos}\left({x}−{A}\right)}{\mathrm{2}}\:=\frac{\mathrm{2}{cosA}}{\mathrm{2}}\:={cosA}. \\ $$
Commented by Rio Michael last updated on 04/Jul/19
$${yeah}. \\ $$
Commented by Rio Michael last updated on 04/Jul/19
$${i}\:{used}\:{the}\:{idea}\:{of}\:{small}\:{angles} \\ $$$${for}\:{small}\:{angles}\:{sin}\theta=\theta \\ $$
Commented by mathmax by abdo last updated on 04/Jul/19
$${you}\:{will}\:{get}\:{the}\:{same}\:{result}. \\ $$
Commented by Rio Michael last updated on 05/Jul/19
$${by}\:{small}\:{angles},\: \\ $$$$\:{A}\left({x}\right)=\:\frac{{sinxcosA}+{cosxsinA}+{sinxcosA}−{cosxsinA}}{\mathrm{2}{x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{2}{sinxcosA}}{\mathrm{2}{x}} \\ $$$$\Rightarrow\underset{{x}\rightarrow\mathrm{0}} {\:{lim}}\frac{\mathrm{2}{sinxcosA}}{\mathrm{2}{x}}\:{for}\:{small}\:{angles}\:{sinx}={x} \\ $$$$\Rightarrow\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{2}{xcosA}}{\mathrm{2}{x}}=\:{cosA}. \\ $$