Menu Close

Prove-lim-x-0-x-sin-x-x-3-1-6-

Tinku Tara 0 Comments
Pin




Question Number 218703 by Nicholas666 last updated on 14/Apr/25
Prove; lim_(x→0) ((x − sin x)/x^3 ) = (1/6)
$$ \\ $$$$\:\:\:\:{Prove};\:{lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{x}\:−\:{sin}\:{x}}{{x}^{\mathrm{3}} }\:=\:\frac{\mathrm{1}}{\mathrm{6}}\:\: \\ $$$$ \\ $$
Answered by SdC355 last updated on 14/Apr/25
sin(x)=Σ (((−)^k )/((2k+1)!))((x/2))^(2k+1) =x−(1/(3!))x^3 +(1/(5!))x^3 ...... x−sin(x)=(1/(3!))x^3 −(1/(5!))x^5 +(1/(7!))x^7 −....... ((x−sin(x))/x^3 )=(1/(3!))−(1/(5!))x^2 +(1/(7!))x^4 −....... lim ′′=(1/6)
$$ \\ $$$$\mathrm{sin}\left({x}\right)=\Sigma\:\frac{\left(−\right)^{{k}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!}\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}{k}+\mathrm{1}} ={x}−\frac{\mathrm{1}}{\mathrm{3}!}{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{5}!}{x}^{\mathrm{3}} …… \\ $$$${x}−\mathrm{sin}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{3}!}{x}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{5}!}{x}^{\mathrm{5}} +\frac{\mathrm{1}}{\mathrm{7}!}{x}^{\mathrm{7}} −……. \\ $$$$\frac{{x}−\mathrm{sin}\left({x}\right)}{{x}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{3}!}−\frac{\mathrm{1}}{\mathrm{5}!}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{7}!}{x}^{\mathrm{4}} −……. \\ $$$$\mathrm{lim}\:''=\frac{\mathrm{1}}{\mathrm{6}} \\ $$
Commented by Nicholas666 last updated on 14/Apr/25
thanks
$${thanks} \\ $$
Answered by Nicholas666 last updated on 14/Apr/25
Answered by aleks041103 last updated on 14/Apr/25
Here is a proof that doesn′t use tailor series. sin(x)=sin(3(x/3))= =sin((x/3))cos(((2x)/3))+sin(((2x)/3))cos((x/3))= =sin((x/3))(1−2sin^2 ((x/3)))+2sin((x/3))cos^2 ((x/3))= =t(1−2t^2 )+2t(1−t^2 )= =t−2t^3 +2t−2t^3 =3t−4t^3 where t=sin(x/3). Let a=lim_(x→0) ((x−sin(x))/x^3 ) (if this limit exists) On the other hand: ((x−sin(x))/x^3 )=((x−3sin(x/3)+4sin^3 (x/3))/x^3 )= =(3/(27)) (((x/3)−sin(x/3))/((x/3)^3 )) + (4/(27)) (((sin(x/3))/((x/3))))^3 We take the limit x→0, which is the same as (x/3)→0 and we get a=(a/9)+(4/(27))⇒9a=a+(4/3)⇒a=(4/(3.8))=(1/6) ⇒lim_(x→0) ((x−sin(x))/x^3 ) = (1/6)
$${Here}\:{is}\:{a}\:{proof}\:{that}\:{doesn}'{t}\:{use}\:{tailor} \\ $$$${series}. \\ $$$$ \\ $$$${sin}\left({x}\right)={sin}\left(\mathrm{3}\frac{{x}}{\mathrm{3}}\right)= \\ $$$$={sin}\left(\frac{{x}}{\mathrm{3}}\right){cos}\left(\frac{\mathrm{2}{x}}{\mathrm{3}}\right)+{sin}\left(\frac{\mathrm{2}{x}}{\mathrm{3}}\right){cos}\left(\frac{{x}}{\mathrm{3}}\right)= \\ $$$$={sin}\left(\frac{{x}}{\mathrm{3}}\right)\left(\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{3}}\right)\right)+\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{3}}\right){cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{3}}\right)= \\ $$$$={t}\left(\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} \right)+\mathrm{2}{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)= \\ $$$$={t}−\mathrm{2}{t}^{\mathrm{3}} +\mathrm{2}{t}−\mathrm{2}{t}^{\mathrm{3}} =\mathrm{3}{t}−\mathrm{4}{t}^{\mathrm{3}} \\ $$$${where}\:{t}={sin}\left({x}/\mathrm{3}\right). \\ $$$$ \\ $$$${Let} \\ $$$${a}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{x}−{sin}\left({x}\right)}{{x}^{\mathrm{3}} }\:\left({if}\:{this}\:{limit}\:{exists}\right) \\ $$$${On}\:{the}\:{other}\:{hand}: \\ $$$$\frac{{x}−{sin}\left({x}\right)}{{x}^{\mathrm{3}} }=\frac{{x}−\mathrm{3}{sin}\left({x}/\mathrm{3}\right)+\mathrm{4}{sin}^{\mathrm{3}} \left({x}/\mathrm{3}\right)}{{x}^{\mathrm{3}} }= \\ $$$$=\frac{\mathrm{3}}{\mathrm{27}}\:\frac{\left({x}/\mathrm{3}\right)−{sin}\left({x}/\mathrm{3}\right)}{\left({x}/\mathrm{3}\right)^{\mathrm{3}} }\:+\:\frac{\mathrm{4}}{\mathrm{27}}\:\left(\frac{{sin}\left({x}/\mathrm{3}\right)}{\left({x}/\mathrm{3}\right)}\right)^{\mathrm{3}} \\ $$$${We}\:{take}\:{the}\:{limit}\:{x}\rightarrow\mathrm{0},\:{which}\:{is}\:{the}\:{same} \\ $$$${as}\:\left({x}/\mathrm{3}\right)\rightarrow\mathrm{0}\:{and}\:{we}\:{get} \\ $$$${a}=\frac{{a}}{\mathrm{9}}+\frac{\mathrm{4}}{\mathrm{27}}\Rightarrow\mathrm{9}{a}={a}+\frac{\mathrm{4}}{\mathrm{3}}\Rightarrow{a}=\frac{\mathrm{4}}{\mathrm{3}.\mathrm{8}}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\Rightarrow\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{{x}−{sin}\left({x}\right)}{{x}^{\mathrm{3}} }\:=\:\frac{\mathrm{1}}{\mathrm{6}} \\ $$
Commented by Nicholas666 last updated on 16/Apr/25
thank you Sir
$${thank}\:{you}\:{Sir} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *