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lim-x-0-x-3-sin-3-x-x-5-




Question Number 101366 by bobhans last updated on 02/Jul/20
lim_(x→0)  ((x^3 −sin^3 x)/x^5 ) =?
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{3}} −\mathrm{sin}\:^{\mathrm{3}} {x}}{{x}^{\mathrm{5}} }\:=? \\ $$
Commented by Dwaipayan Shikari last updated on 02/Jul/20
lim_(x→0) ((x^3 −(x−(x^3 /6))^3 )/x^5 )=lim_(x→0) ((x^3 −x^3 +(x^9 /6)+((3x^5 )/6)−((3x^7 )/(36)))/x^5 )=(((3x^5 )/6)/x^5 )=(1/2)  {(x^9 /6) and ((3x^7 )/(36))are  negligable}
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}^{\mathrm{3}} −\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\right)^{\mathrm{3}} }{{x}^{\mathrm{5}} }=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}^{\mathrm{3}} −{x}^{\mathrm{3}} +\frac{{x}^{\mathrm{9}} }{\mathrm{6}}+\frac{\mathrm{3}{x}^{\mathrm{5}} }{\mathrm{6}}−\frac{\mathrm{3}{x}^{\mathrm{7}} }{\mathrm{36}}}{{x}^{\mathrm{5}} }=\frac{\frac{\mathrm{3}{x}^{\mathrm{5}} }{\mathrm{6}}}{{x}^{\mathrm{5}} }=\frac{\mathrm{1}}{\mathrm{2}}\:\:\left\{\frac{{x}^{\mathrm{9}} }{\mathrm{6}}\:{and}\:\frac{\mathrm{3}{x}^{\mathrm{7}} }{\mathrm{36}}{are}\:\:{negligable}\right\} \\ $$
Commented by bemath last updated on 02/Jul/20
agree
$$\mathrm{agree} \\ $$
Answered by ajfour last updated on 02/Jul/20
lim_(x→0) (((x−sin x)(x^2 +xsin x+sin^2 x))/x^5 )  lim_(x→0) (((x−sin x))/x^3 )×lim_(x→0) (((x^2 +xsin x+sin^2 x))/x^2 )  =lim_(x→0) ((x−(x−(x^3 /(3!))+....))/x^3 )×(3)  =  (1/(3!))×3 = (1/2) .
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({x}−\mathrm{sin}\:{x}\right)\left({x}^{\mathrm{2}} +{x}\mathrm{sin}\:{x}+\mathrm{sin}\:^{\mathrm{2}} {x}\right)}{{x}^{\mathrm{5}} } \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({x}−\mathrm{sin}\:{x}\right)}{{x}^{\mathrm{3}} }×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({x}^{\mathrm{2}} +{x}\mathrm{sin}\:{x}+\mathrm{sin}\:^{\mathrm{2}} {x}\right)}{{x}^{\mathrm{2}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}−\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+….\right)}{{x}^{\mathrm{3}} }×\left(\mathrm{3}\right) \\ $$$$=\:\:\frac{\mathrm{1}}{\mathrm{3}!}×\mathrm{3}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:. \\ $$
Commented by bemath last updated on 02/Jul/20
yes..thank you sir
$$\mathrm{yes}..\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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