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Question Number 83587 by jagoll last updated on 04/Mar/20
lim_(x→0) ((3sin πx−sin 3πx)/x^3 )
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3sin}\:\pi\mathrm{x}−\mathrm{sin}\:\mathrm{3}\pi\mathrm{x}}{\mathrm{x}^{\mathrm{3}} } \\ $$
Commented by john santu last updated on 04/Mar/20
let u = πx ⇒ x= (u/π) lim_(u→0) ((3sin u−sin 3u)/u^3 ) × π^3 = π^3 × lim_(u→0) ((3sin u−(3sin u−4sin^3 u))/u^3 ) π^3 × lim_(u→0) ((4sin^3 u)/u^3 ) = 4π^3
$$\mathrm{let}\:\mathrm{u}\:=\:\pi\mathrm{x}\:\Rightarrow\:\mathrm{x}=\:\frac{\mathrm{u}}{\pi} \\ $$$$\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3sin}\:\mathrm{u}−\mathrm{sin}\:\mathrm{3u}}{\mathrm{u}^{\mathrm{3}} }\:×\:\pi^{\mathrm{3}} \:=\: \\ $$$$\pi^{\mathrm{3}} ×\:\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3sin}\:\mathrm{u}−\left(\mathrm{3sin}\:\mathrm{u}−\mathrm{4sin}\:^{\mathrm{3}} \mathrm{u}\right)}{\mathrm{u}^{\mathrm{3}} } \\ $$$$\pi^{\mathrm{3}} \:×\:\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{4sin}\:^{\mathrm{3}} \mathrm{u}}{\mathrm{u}^{\mathrm{3}} }\:=\:\mathrm{4}\pi^{\mathrm{3}} \\ $$
Commented by mathmax by abdo last updated on 04/Mar/20
we have sinu =Σ_(n=0) ^∞ (((−1)^n )/((2n+1)!))u^(2n+1) =u−(u^3 /(3!))+... ⇒ sin(πx)=πx−(((πx)^3 )/(3!)) +o(x^5 ) sin(3πx) =3πx−(((3πx)^3 )/(3!)) +o^′ (x^5 ) ⇒3sin(πx)−sin(3πx) =3πx −(1/2)π^3 x^3 −3πx +(9/2)π^3 x^3 +o(x^5 ) =4π^3 x^3 +o(x^5 ) ⇒ ((3sin(πx)−sin(3πx))/x^3 ) =4π^3 +o(x^2 ) ⇒ lim_(x→0) ((3sin(πx)−sin(3πx))/x^3 ) =4π^3
$${we}\:{have}\:{sinu}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{u}^{\mathrm{2}{n}+\mathrm{1}} \:\:={u}−\frac{{u}^{\mathrm{3}} }{\mathrm{3}!}+…\:\Rightarrow \\ $$$${sin}\left(\pi{x}\right)=\pi{x}−\frac{\left(\pi{x}\right)^{\mathrm{3}} }{\mathrm{3}!}\:+{o}\left({x}^{\mathrm{5}} \right) \\ $$$${sin}\left(\mathrm{3}\pi{x}\right)\:=\mathrm{3}\pi{x}−\frac{\left(\mathrm{3}\pi{x}\right)^{\mathrm{3}} }{\mathrm{3}!}\:+{o}^{'} \left({x}^{\mathrm{5}} \right)\:\Rightarrow\mathrm{3}{sin}\left(\pi{x}\right)−{sin}\left(\mathrm{3}\pi{x}\right) \\ $$$$=\mathrm{3}\pi{x}\:−\frac{\mathrm{1}}{\mathrm{2}}\pi^{\mathrm{3}} \:{x}^{\mathrm{3}} −\mathrm{3}\pi{x}\:+\frac{\mathrm{9}}{\mathrm{2}}\pi^{\mathrm{3}} {x}^{\mathrm{3}} \:+{o}\left({x}^{\mathrm{5}} \right)\:=\mathrm{4}\pi^{\mathrm{3}} \:{x}^{\mathrm{3}} \:+{o}\left({x}^{\mathrm{5}} \right)\:\Rightarrow \\ $$$$\frac{\mathrm{3}{sin}\left(\pi{x}\right)−{sin}\left(\mathrm{3}\pi{x}\right)}{{x}^{\mathrm{3}} }\:=\mathrm{4}\pi^{\mathrm{3}} \:+{o}\left({x}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\frac{\mathrm{3}{sin}\left(\pi{x}\right)−{sin}\left(\mathrm{3}\pi{x}\right)}{{x}^{\mathrm{3}} }\:=\mathrm{4}\pi^{\mathrm{3}} \\ $$

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