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Question Number 110449 by mathmax by abdo last updated on 29/Aug/20

find lim_(x→0) ((arctan(x−sinx)−arctan(1−cosx))/x^2 )

$$\mathrm{find}\:\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{arctan}\left(\mathrm{x}−\mathrm{sinx}\right)−\mathrm{arctan}\left(\mathrm{1}−\mathrm{cosx}\right)}{\mathrm{x}^{\mathrm{2}} } \\ $$

Answered by bemath last updated on 29/Aug/20

lim_(x→0) ((x−sin x−(1−cos x))/x^2 ) = lim_(x→0) ((x−(x−(1/6)x^3 )−(1−(1−(1/2)x^2 )))/x^2 ) = lim_(x→0) (((1/6)x^3 −(1/2)x^2 )/x^2 ) = −(1/2)

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{x}−\mathrm{sin}\:\mathrm{x}−\left(\mathrm{1}−\mathrm{cos}\:\mathrm{x}\right)}{\mathrm{x}^{\mathrm{2}} }\:= \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{x}−\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{6}}\mathrm{x}^{\mathrm{3}} \right)−\left(\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} \right)\right)}{\mathrm{x}^{\mathrm{2}} }\:= \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{1}}{\mathrm{6}}\mathrm{x}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} }\:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by mathmax by abdo last updated on 29/Aug/20

let use hospital theorem u(x)=arctan(x−sinx)−arctan(1−cosx) v(x) =x^2 we have u^′ (x) =((1−cosx)/(1+(x−sinx)^2 ))−((sinx)/(1+(1−cosx)^2 )) ⇒u^((2)) (x)=((sinx(1+(x−sinx)^2 )−2(1−cosx)(1−cosx))/({1+(x−sinx}^2 )) −((cosx{1+(1−cosx)^2 }−2{sinx}{1+(1−cosx)^2 })/({1+(1−cosx)^2 }^2 )) ⇒lim_(x→0) u^((2)) (x) =0−(1/1) =−1 also v^′ (x)=2x ⇒v^((2)) (x)=2 ⇒ lim_(x→0) v^((2)) (x) =2 ⇒lim_(x→0) ((u^((2)) (x))/(v^((2)) (x))) =−(1/2)

$$\mathrm{let}\:\mathrm{use}\:\mathrm{hospital}\:\mathrm{theorem}\:\:\:\mathrm{u}\left(\mathrm{x}\right)=\mathrm{arctan}\left(\mathrm{x}−\mathrm{sinx}\right)−\mathrm{arctan}\left(\mathrm{1}−\mathrm{cosx}\right) \\ $$$$\mathrm{v}\left(\mathrm{x}\right)\:=\mathrm{x}^{\mathrm{2}} \:\:\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{u}^{'} \left(\mathrm{x}\right)\:=\frac{\mathrm{1}−\mathrm{cosx}}{\mathrm{1}+\left(\mathrm{x}−\mathrm{sinx}\right)^{\mathrm{2}} }−\frac{\mathrm{sinx}}{\mathrm{1}+\left(\mathrm{1}−\mathrm{cosx}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{u}^{\left(\mathrm{2}\right)} \left(\mathrm{x}\right)=\frac{\mathrm{sinx}\left(\mathrm{1}+\left(\mathrm{x}−\mathrm{sinx}\right)^{\mathrm{2}} \right)−\mathrm{2}\left(\mathrm{1}−\mathrm{cosx}\right)\left(\mathrm{1}−\mathrm{cosx}\right)}{\left\{\mathrm{1}+\left(\mathrm{x}−\mathrm{sinx}\right\}^{\mathrm{2}} \right.} \\ $$$$−\frac{\mathrm{cosx}\left\{\mathrm{1}+\left(\mathrm{1}−\mathrm{cosx}\right)^{\mathrm{2}} \right\}−\mathrm{2}\left\{\mathrm{sinx}\right\}\left\{\mathrm{1}+\left(\mathrm{1}−\mathrm{cosx}\right)^{\mathrm{2}} \right\}}{\left\{\mathrm{1}+\left(\mathrm{1}−\mathrm{cosx}\right)^{\mathrm{2}} \right\}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{u}^{\left(\mathrm{2}\right)} \left(\mathrm{x}\right)\:=\mathrm{0}−\frac{\mathrm{1}}{\mathrm{1}}\:=−\mathrm{1}\:\:\mathrm{also}\:\mathrm{v}^{'} \left(\mathrm{x}\right)=\mathrm{2x}\:\Rightarrow\mathrm{v}^{\left(\mathrm{2}\right)} \left(\mathrm{x}\right)=\mathrm{2}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{v}^{\left(\mathrm{2}\right)} \left(\mathrm{x}\right)\:=\mathrm{2}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\:\frac{\mathrm{u}^{\left(\mathrm{2}\right)} \left(\mathrm{x}\right)}{\mathrm{v}^{\left(\mathrm{2}\right)} \left(\mathrm{x}\right)}\:=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$

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