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lim-x-0-x-sinx-x-3-solve-without-hopital-and-any-series-

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Question Number 191078 by sciencestudentW last updated on 17/Apr/23
lim_(x→0) ((x−sinx)/x^3 )=? solve without hopital and any series.
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}−{sinx}}{{x}^{\mathrm{3}} }=? \\ $$$${solve}\:{without}\:{hopital}\:{and}\:{any}\:{series}. \\ $$
Answered by mehdee42 last updated on 17/Apr/23
if x→0 ⇒x−sinx ∼ (1/6)x^3
$${if}\:{x}\rightarrow\mathrm{0}\:\:\Rightarrow{x}−{sinx}\:\sim\:\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} \\ $$
Answered by mathlove last updated on 18/Apr/23
l=lim_(x→0) ((x−sinx)/x^3 ) lim_(x→0) ((x−2sin(x/2)cos(x/2))/x^3 )=2lim_(x→0) (((x/2)−sin(x/2)cos(x/2)−sin(x/2)+sin(x/2))/x^3 ) 2lim_(x→0) (((x/2)−sin(x/2))/x^3 )+2lim_(x→0) ((sin(x/2)−sin(x/2)cos(x/2))/x^3 ) 2lim_(x→0) (((x/2)−sin(x/2))/(8((x/2))^3 ))+2lim_(x→0) ((sin(x/2)(1−cos(x/2)))/(8((x/2))((x/2))^2 )) (1/4)lim_(x→0) (((x/2)−sin(x/2))/(((x/2))^3 ))+(1/4)lim_(x→0) ((sin(x/2))/(x/2))lim_(x→0) ((1−cos(x/2))/(((x/2))^2 )) (1/4)l+(1/4)×(1/2)⇒l−(1/4)l=(1/8) l=(1/6) lim_(x→0) ((x−sinx)/x^3 )=(1/6)
$${l}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}−{sinx}}{{x}^{\mathrm{3}} } \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}−\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}}{{x}^{\mathrm{3}} }=\mathrm{2}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{{x}}{\mathrm{2}}−{sin}\frac{{x}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}−{sin}\frac{{x}}{\mathrm{2}}+{sin}\frac{{x}}{\mathrm{2}}}{{x}^{\mathrm{3}} } \\ $$$$\mathrm{2}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{{x}}{\mathrm{2}}−{sin}\frac{{x}}{\mathrm{2}}}{{x}^{\mathrm{3}} }+\mathrm{2}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{sin}\frac{{x}}{\mathrm{2}}−{sin}\frac{{x}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}}{{x}^{\mathrm{3}} } \\ $$$$\mathrm{2}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{{x}}{\mathrm{2}}−{sin}\frac{{x}}{\mathrm{2}}}{\mathrm{8}\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{3}} }+\mathrm{2}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{sin}\frac{{x}}{\mathrm{2}}\left(\mathrm{1}−{cos}\frac{{x}}{\mathrm{2}}\right)}{\mathrm{8}\left(\frac{{x}}{\mathrm{2}}\right)\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{{x}}{\mathrm{2}}−{sin}\frac{{x}}{\mathrm{2}}}{\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{4}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{sin}\frac{{x}}{\mathrm{2}}}{\frac{{x}}{\mathrm{2}}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−{cos}\frac{{x}}{\mathrm{2}}}{\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}{l}+\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{l}−\frac{\mathrm{1}}{\mathrm{4}}{l}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${l}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}−{sinx}}{{x}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{6}} \\ $$

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