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Question Number 112917 by abdullahquwatan last updated on 10/Sep/20

lim_(x→∞) [csc^2 ((2/x))−(1/4)x^2 ]

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left[\mathrm{csc}^{\mathrm{2}} \left(\frac{\mathrm{2}}{\mathrm{x}}\right)−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{x}^{\mathrm{2}} \right] \\ $$

Commented by bobhans last updated on 10/Sep/20

lim_(w→0) (((2w+sin (2w))/(4w)))(((2w−sin (2w))/(w.sin^2 (2w)))) 1×lim_(w→0) (((2w−sin 2w)/(4w^3 )))= lim_(w→0) ((2−2cos 2w)/(12w^2 )) = lim_(w→0) (((1−cos 2w)/(6w^2 ))) =lim_(w→0) (((2sin^2 w)/(6w^2 )))=(1/3)

$$\underset{\mathrm{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{2w}+\mathrm{sin}\:\left(\mathrm{2w}\right)}{\mathrm{4w}}\right)\left(\frac{\mathrm{2w}−\mathrm{sin}\:\left(\mathrm{2w}\right)}{\mathrm{w}.\mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{2w}\right)}\right) \\ $$$$\mathrm{1}×\underset{\mathrm{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{2w}−\mathrm{sin}\:\mathrm{2w}}{\mathrm{4w}^{\mathrm{3}} }\right)= \\ $$$$\underset{\mathrm{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}−\mathrm{2cos}\:\mathrm{2w}}{\mathrm{12w}^{\mathrm{2}} }\:=\:\underset{\mathrm{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2w}}{\mathrm{6w}^{\mathrm{2}} }\right) \\ $$$$=\underset{\mathrm{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{2sin}\:^{\mathrm{2}} \mathrm{w}}{\mathrm{6w}^{\mathrm{2}} }\right)=\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Commented by abdullahquwatan last updated on 10/Sep/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by abdullahquwatan last updated on 10/Sep/20

why 4w^3 ??

$$\mathrm{why}\:\mathrm{4w}^{\mathrm{3}} \:?? \\ $$

Answered by mathmax by abdo last updated on 10/Sep/20

let f(x) =(1/(sin^2 ((2/x))))−(x^2 /4) changement (2/x)=t give (x/2)=(1/t) ⇒x=(2/t) (x→∞ ⇔t→0) ⇒f(x) =(1/(sin^2 t))−(1/4)((4/t^2 )) =(1/(sin^2 t))−(1/t^2 )=g(t) =((t^2 −sin^2 t)/(t^2 sin^2 t)) we have sint ∼t−(t^3 /6) ⇒sin^2 t ∼(t−(t^3 /6))^2 =t^2 −(1/3)t^4 +(t^6 /(36)) ∼t^2 −(1/3)t^4 ⇒g(t) ∼((t^2 −t^2 +(1/3)t^4 )/(t^2 .t^2 )) =(1/3) ⇒lim_(t→0) g(t)=(1/3) ⇒ lim_(x→∞) f(x) =(1/3)

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \left(\frac{\mathrm{2}}{\mathrm{x}}\right)}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{4}}\:\:\mathrm{changement}\:\frac{\mathrm{2}}{\mathrm{x}}=\mathrm{t}\:\mathrm{give}\:\frac{\mathrm{x}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{t}}\:\Rightarrow\mathrm{x}=\frac{\mathrm{2}}{\mathrm{t}} \\ $$$$\left(\mathrm{x}\rightarrow\infty\:\Leftrightarrow\mathrm{t}\rightarrow\mathrm{0}\right)\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \mathrm{t}}−\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{4}}{\mathrm{t}^{\mathrm{2}} }\right)\:=\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \mathrm{t}}−\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }=\mathrm{g}\left(\mathrm{t}\right) \\ $$$$=\frac{\mathrm{t}^{\mathrm{2}} −\mathrm{sin}^{\mathrm{2}} \mathrm{t}}{\mathrm{t}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \mathrm{t}}\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{sint}\:\sim\mathrm{t}−\frac{\mathrm{t}^{\mathrm{3}} }{\mathrm{6}}\:\Rightarrow\mathrm{sin}^{\mathrm{2}} \mathrm{t}\:\sim\left(\mathrm{t}−\frac{\mathrm{t}^{\mathrm{3}} }{\mathrm{6}}\right)^{\mathrm{2}} \:=\mathrm{t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{3}}\mathrm{t}^{\mathrm{4}} +\frac{\mathrm{t}^{\mathrm{6}} }{\mathrm{36}} \\ $$$$\sim\mathrm{t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{3}}\mathrm{t}^{\mathrm{4}} \:\Rightarrow\mathrm{g}\left(\mathrm{t}\right)\:\sim\frac{\mathrm{t}^{\mathrm{2}} −\mathrm{t}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{t}^{\mathrm{4}} }{\mathrm{t}^{\mathrm{2}} .\mathrm{t}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\mathrm{lim}_{\mathrm{t}\rightarrow\mathrm{0}} \mathrm{g}\left(\mathrm{t}\right)=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\infty} \mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Commented by abdullahquwatan last updated on 10/Sep/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by mathmax by abdo last updated on 11/Sep/20

you are welcome

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$

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