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Question Number 110760 by bobhans last updated on 30/Aug/20

Answered by john santu last updated on 30/Aug/20

{ ((x+1=m)),((m→0)) :} L=lim_(m→0) ((sin 2m−tan m)/m^3 ) L= lim_(m→0) ((2sin m cos m−tan m)/m^3 ) L=lim_(m→0) ((tan m(2cos^2 m−1))/m^3 ) L=lim_(m→0) ((2cos^2 m−1)/m^2 ) = ∞

$$\begin{cases}{{x}+\mathrm{1}={m}}\\{{m}\rightarrow\mathrm{0}}\end{cases} \\ $$$${L}=\underset{{m}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{2}{m}−\mathrm{tan}\:{m}}{{m}^{\mathrm{3}} } \\ $$$${L}=\:\underset{{m}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:{m}\:\mathrm{cos}\:{m}−\mathrm{tan}\:{m}}{{m}^{\mathrm{3}} } \\ $$$${L}=\underset{{m}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{tan}\:{m}\left(\mathrm{2cos}\:^{\mathrm{2}} {m}−\mathrm{1}\right)}{{m}^{\mathrm{3}} } \\ $$$${L}=\underset{{m}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2cos}\:^{\mathrm{2}} {m}−\mathrm{1}}{{m}^{\mathrm{2}} }\:=\:\infty \\ $$

Answered by mathmax by abdo last updated on 30/Aug/20

let f(x) =((sin(2x+2)−tan(x+1))/((x+1)^3 )) changement x+1 =t give f(x)=((sin(2t)−tan(t))/t^3 ) =g(t) (x→−1⇒t→0) we have sin(2t) ∼2t−(((2t)^3 )/6) +o(t^3 ) =2t−(4/3) t^3 +o(t^3 ) tant ∼ t+(t^3 /3) +o(t^3 ) ⇒g(t)∼((2t−(4/3)t^3 −t−(t^3 /3))/t^3 ) =(1/t^2 )−(5/3) ⇒lim_(t→0) g(t) =+∞ =lim_(x→−1) f(x)

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{sin}\left(\mathrm{2x}+\mathrm{2}\right)−\mathrm{tan}\left(\mathrm{x}+\mathrm{1}\right)}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}} }\:\:\mathrm{changement}\:\mathrm{x}+\mathrm{1}\:=\mathrm{t}\:\mathrm{give} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{sin}\left(\mathrm{2t}\right)−\mathrm{tan}\left(\mathrm{t}\right)}{\mathrm{t}^{\mathrm{3}} }\:=\mathrm{g}\left(\mathrm{t}\right)\:\:\:\:\:\left(\mathrm{x}\rightarrow−\mathrm{1}\Rightarrow\mathrm{t}\rightarrow\mathrm{0}\right) \\ $$$$\mathrm{we}\:\mathrm{have}\:\:\mathrm{sin}\left(\mathrm{2t}\right)\:\sim\mathrm{2t}−\frac{\left(\mathrm{2t}\right)^{\mathrm{3}} }{\mathrm{6}}\:+\mathrm{o}\left(\mathrm{t}^{\mathrm{3}} \right)\:=\mathrm{2t}−\frac{\mathrm{4}}{\mathrm{3}}\:\mathrm{t}^{\mathrm{3}} \:+\mathrm{o}\left(\mathrm{t}^{\mathrm{3}} \right) \\ $$$$\mathrm{tant}\:\sim\:\mathrm{t}+\frac{\mathrm{t}^{\mathrm{3}} }{\mathrm{3}}\:+\mathrm{o}\left(\mathrm{t}^{\mathrm{3}} \right)\:\Rightarrow\mathrm{g}\left(\mathrm{t}\right)\sim\frac{\mathrm{2t}−\frac{\mathrm{4}}{\mathrm{3}}\mathrm{t}^{\mathrm{3}} −\mathrm{t}−\frac{\mathrm{t}^{\mathrm{3}} }{\mathrm{3}}}{\mathrm{t}^{\mathrm{3}} }\:=\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }−\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{lim}_{\mathrm{t}\rightarrow\mathrm{0}} \mathrm{g}\left(\mathrm{t}\right)\:=+\infty\:=\mathrm{lim}_{\mathrm{x}\rightarrow−\mathrm{1}} \mathrm{f}\left(\mathrm{x}\right) \\ $$

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