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Question Number 84460 by jagoll last updated on 13/Mar/20
lim_(x→0)  ((sin (2+x)−sin (2−x))/x)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(\mathrm{2}+\mathrm{x}\right)−\mathrm{sin}\:\left(\mathrm{2}−\mathrm{x}\right)}{\mathrm{x}} \\ $$
Commented by jagoll last updated on 13/Mar/20
lim_(x→0)  ((cos (2+x)+cos (2−x))/1)  = 2cos 2
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left(\mathrm{2}+\mathrm{x}\right)+\mathrm{cos}\:\left(\mathrm{2}−\mathrm{x}\right)}{\mathrm{1}} \\ $$$$=\:\mathrm{2cos}\:\mathrm{2} \\ $$
Commented by mathmax by abdo last updated on 13/Mar/20
let f(x)=((sin(2+x)−sin(2−x))/x) ⇒  f(x) =((sin2 cosx +cos2sinx−(sin2 cosx −cos2 sinx))/x)  =((2cos2 sinx)/x) ∼ 2cos2    (x→0) ⇒lim_(x→0) f(x)=2cos2
$${let}\:{f}\left({x}\right)=\frac{{sin}\left(\mathrm{2}+{x}\right)−{sin}\left(\mathrm{2}−{x}\right)}{{x}}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\frac{{sin}\mathrm{2}\:{cosx}\:+{cos}\mathrm{2}{sinx}−\left({sin}\mathrm{2}\:{cosx}\:−{cos}\mathrm{2}\:{sinx}\right)}{{x}} \\ $$$$=\frac{\mathrm{2}{cos}\mathrm{2}\:{sinx}}{{x}}\:\sim\:\mathrm{2}{cos}\mathrm{2}\:\:\:\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} {f}\left({x}\right)=\mathrm{2}{cos}\mathrm{2} \\ $$

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