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Question Number 66696 by mathmax by abdo last updated on 18/Aug/19

calculate lim_(x→0) ((arctan(1+x^3 )−(π/4))/(xsin(x^2 )))

$${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\:\frac{{arctan}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)−\frac{\pi}{\mathrm{4}}}{{xsin}\left({x}^{\mathrm{2}} \right)} \\ $$

Commented by kaivan.ahmadi last updated on 18/Aug/19

lim_(x→0) ((3x^2 /(1+(1+x^3 )^2 ))/(3x^2 ))=lim_(x→0) (1/1+(1+x^3 )^2 )=1/2

$${lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{3}{x}^{\mathrm{2}} /\left(\mathrm{1}+\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{\mathrm{2}} \right)}{\mathrm{3}{x}^{\mathrm{2}} }={lim}_{{x}\rightarrow\mathrm{0}} \left(\mathrm{1}/\mathrm{1}+\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{\mathrm{2}} \right)=\mathrm{1}/\mathrm{2} \\ $$

Commented by ~ À ® @ 237 ~ last updated on 18/Aug/19

let named it L L=lim_(x→0) ((arctan(1+x^3 ) −(π/4))/x^3 ) . (x^2 /(sin(x^2 ))) Now lim_(x→0) (x^2 /(sin(x^2 ))) = lim_(u→0) (1/((sinu)/u)) = 1 ( u=x^2 ) lim_(x→0) ((arctan(1+x^3 )−(π/4))/x^3 ) = lim_(v→0) ((arctan(1+v)−(π/4))/v) =lim_(v→0) (1/(1+(1+v)^2 )) = (1/2) (v=x^3 ) So L=(1/2)

$${let}\:{named}\:{it}\:{L} \\ $$$${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{arctan}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)\:−\frac{\pi}{\mathrm{4}}}{{x}^{\mathrm{3}} }\:.\:\frac{{x}^{\mathrm{2}} }{{sin}\left({x}^{\mathrm{2}} \right)}\: \\ $$$${Now}\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} }{{sin}\left({x}^{\mathrm{2}} \right)}\:=\:\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\frac{{sinu}}{{u}}}\:=\:\mathrm{1}\:\:\:\:\:\:\:\:\left(\:{u}={x}^{\mathrm{2}} \right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{arctan}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)−\frac{\pi}{\mathrm{4}}}{{x}^{\mathrm{3}} }\:=\:\underset{{v}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{arctan}\left(\mathrm{1}+{v}\right)−\frac{\pi}{\mathrm{4}}}{{v}}\:=\underset{{v}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{1}+{v}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\left({v}={x}^{\mathrm{3}} \right) \\ $$$${So}\:\:\:\:{L}=\frac{\mathrm{1}}{\mathrm{2}}\: \\ $$$$ \\ $$$$ \\ $$

Commented by mathmax by abdo last updated on 19/Aug/19

let f(u) =arctan(1+u) we have f(u)=f(0)+u f^′ (0)+o(u^2 ) but f(0) =(π/4) f^′ (u) =(1/(1+(1+u)^2 )) ⇒f^′ (0) =(1/2) ⇒f(u) =(π/4) +(1/2)u +o(u^2 ) ⇒ arctan(1+x^3 ) =(π/4) +(x^3 /2) +o(x^6 ) ⇒arctan(1+x^3 )−(π/4)∼(x^3 /2) (x→0) slso xsin(x^2 ) ∼ x^3 ⇒ lim_(x→0) ((arctan(1+x^3 )−(π/4))/(xsin(x^2 ))) =lim_(x→0) (x^3 /(2x^3 )) =(1/2) .

$${let}\:{f}\left({u}\right)\:={arctan}\left(\mathrm{1}+{u}\right)\:\:{we}\:{have}\: \\ $$$${f}\left({u}\right)={f}\left(\mathrm{0}\right)+{u}\:{f}^{'} \left(\mathrm{0}\right)+{o}\left({u}^{\mathrm{2}} \right)\:\:{but}\:{f}\left(\mathrm{0}\right)\:=\frac{\pi}{\mathrm{4}} \\ $$$${f}^{'} \left({u}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{1}+{u}\right)^{\mathrm{2}} }\:\Rightarrow{f}^{'} \left(\mathrm{0}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{f}\left({u}\right)\:=\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{2}}{u}\:+{o}\left({u}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$${arctan}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)\:=\frac{\pi}{\mathrm{4}}\:+\frac{{x}^{\mathrm{3}} }{\mathrm{2}}\:+{o}\left({x}^{\mathrm{6}} \right)\:\Rightarrow{arctan}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)−\frac{\pi}{\mathrm{4}}\sim\frac{{x}^{\mathrm{3}} }{\mathrm{2}}\:\:\left({x}\rightarrow\mathrm{0}\right) \\ $$$${slso}\:{xsin}\left({x}^{\mathrm{2}} \right)\:\sim\:{x}^{\mathrm{3}} \:\Rightarrow\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{{arctan}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)−\frac{\pi}{\mathrm{4}}}{{xsin}\left({x}^{\mathrm{2}} \right)}\:={lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{x}^{\mathrm{3}} }{\mathrm{2}{x}^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:. \\ $$

Answered by Cmr 237 last updated on 18/Aug/19

Answered by Cmr 237 last updated on 18/Aug/19

with the limited develpment we have: arctan(1+x^3 )=(π/4)+(x^3 /2) +o(x^3 ) sin(x^2 )=x^2 +o(x^2 ) so lim _(x→0) ((arctan(1+x^3 )−(π/4))/(xsin(x^2 )))=lim_ _(x→0) (((π/4)+(x^3 /2)−(π/4))/(x×x^2 )) =(1/2).

$${with}\:{the}\:{limited}\:{develpment}\:{we}\: \\ $$$${have}: \\ $$$${arctan}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)=\frac{\pi}{\mathrm{4}}+\frac{{x}^{\mathrm{3}} }{\mathrm{2}}\:+{o}\left({x}^{\mathrm{3}} \right) \\ $$$${sin}\left({x}^{\mathrm{2}} \right)={x}^{\mathrm{2}} +{o}\left({x}^{\mathrm{2}} \right) \\ $$$${so} \\ $$$${lim}\underset{{x}\rightarrow\mathrm{0}} {\:}\frac{{arctan}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)−\frac{\pi}{\mathrm{4}}}{{xsin}\left({x}^{\mathrm{2}} \right)}={li}\underset{} {{m}}\underset{{x}\rightarrow\mathrm{0}} {\:}\frac{\frac{\pi}{\mathrm{4}}+\frac{{x}^{\mathrm{3}} }{\mathrm{2}}−\frac{\pi}{\mathrm{4}}}{{x}×{x}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}. \\ $$

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