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Question Number 69607 by malwaan last updated on 25/Sep/19
without using lhospital please prove that lim_(x→0) ((x−sin x)/x^3 ) = (1/6) I want every method possible because someone challenge me
$$\boldsymbol{{without}}\:\boldsymbol{{using}}\:\boldsymbol{{lhospital}}\:\boldsymbol{{please}} \\ $$$$\boldsymbol{{prove}}\:\boldsymbol{{that}} \\ $$$$\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\:\frac{\boldsymbol{{x}}−\boldsymbol{{sin}}\:\boldsymbol{{x}}}{\boldsymbol{{x}}^{\mathrm{3}} }\:=\:\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\boldsymbol{{I}}\:\boldsymbol{{want}}\:\boldsymbol{{every}}\:\boldsymbol{{method}} \\ $$$$\boldsymbol{{possible}}\:\boldsymbol{{because}}\:\boldsymbol{{someone}} \\ $$$$\boldsymbol{{challenge}}\:\boldsymbol{{me}}\: \\ $$
Commented by mathmax by abdo last updated on 26/Sep/19
we have sinx =Σ_(n=0) ^∞ (((−1)^n x^(2n+1) )/((2n+1)!)) =x−(x^3 /(3!)) +(x^5 /(5!)) −(x^7 /(7!)) +... ⇒ −sinx =−x+(x^3 /(3!))−(x^5 /(5!)) +(x^7 /(7!)) +o(x^7 ) ⇒x−sinx = (x^3 /(3!))−(x^5 /(5!)) +(x^7 /(7!)) +o(x^7 ) ⇒((x−sinx)/x^3 ) =(1/(3!))−(x^2 /(5!)) +(x^4 /(7!)) +o(x^4 ) ⇒ lim_(x→0) ((x−sinx)/x^3 ) =(1/(3!)) =(1/6)
$${we}\:{have}\:\:{sinx}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}\:+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}\:−\frac{{x}^{\mathrm{7}} }{\mathrm{7}!}\:+…\:\Rightarrow \\ $$$$−{sinx}\:=−{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}−\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}\:+\frac{{x}^{\mathrm{7}} }{\mathrm{7}!}\:+{o}\left({x}^{\mathrm{7}} \right)\:\Rightarrow{x}−{sinx}\:= \\ $$$$\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}−\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}\:+\frac{{x}^{\mathrm{7}} }{\mathrm{7}!}\:+{o}\left({x}^{\mathrm{7}} \right)\:\Rightarrow\frac{{x}−{sinx}}{{x}^{\mathrm{3}} }\:=\frac{\mathrm{1}}{\mathrm{3}!}−\frac{{x}^{\mathrm{2}} }{\mathrm{5}!}\:+\frac{{x}^{\mathrm{4}} }{\mathrm{7}!}\:+{o}\left({x}^{\mathrm{4}} \right)\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{{x}−{sinx}}{{x}^{\mathrm{3}} }\:=\frac{\mathrm{1}}{\mathrm{3}!}\:=\frac{\mathrm{1}}{\mathrm{6}} \\ $$
Commented by Prithwish sen last updated on 25/Sep/19
lim_(x→0) ((x−(x−(x^3 /(3!))+(x^5 /(5!))−....))/x^3 ) = lim_(x→0) (1/(3!)) − (x^2 /(5!)) + the higher power of x = (1/(3!)) = (1/6) proved
$$\mathrm{li}\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{m}}\:\frac{\mathrm{x}−\left(\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{\mathrm{x}^{\mathrm{5}} }{\mathrm{5}!}−….\right)}{\mathrm{x}^{\mathrm{3}} }\:=\:\mathrm{li}\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{m}}\:\:\frac{\mathrm{1}}{\mathrm{3}!}\:−\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{5}!}\:+\:\mathrm{the}\:\mathrm{higher}\:\mathrm{power}\:\mathrm{of}\:\mathrm{x} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}!}\:=\:\frac{\mathrm{1}}{\mathrm{6}}\:\:\boldsymbol{\mathrm{proved}} \\ $$
Commented by malwaan last updated on 26/Sep/19
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Commented by malwaan last updated on 27/Sep/19
thank you so much
$${thank}\:{you}\:{so}\:{much} \\ $$

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