Menu Close

lim-x-pi-2-2-cos-x-1-x-x-pi-2-

Tinku Tara 0 Comments
Pin




Question Number 131769 by liberty last updated on 08/Feb/21
lim_(x→π/2) ((2^(−cos x) −1)/(x(x−(π/2)))) =?
$$\:\:\:\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\frac{\mathrm{2}^{−\mathrm{cos}\:\mathrm{x}} \:−\mathrm{1}}{\mathrm{x}\left(\mathrm{x}−\frac{\pi}{\mathrm{2}}\right)}\:=?\: \\ $$
Answered by bemath last updated on 08/Feb/21
lim_(x→π/2) (1/x) . lim_(x→π/2) ((2^(sin (x−(π/2))) −1)/(x−(π/2))) = (2/π) . lim_(h→0) ((2^(sin h) −1)/h) = (2/π) . lim_(h→0) ((ln 2. 2^(sin h) .cos h)/1) = ((2ln 2)/π) = π^(−1) .ln 4
$$\:\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{x}}\:.\:\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{2}^{\mathrm{sin}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{2}}\right)} −\mathrm{1}}{\mathrm{x}−\frac{\pi}{\mathrm{2}}}\:= \\ $$$$\:\frac{\mathrm{2}}{\pi}\:.\:\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}^{\mathrm{sin}\:\mathrm{h}} −\mathrm{1}}{\mathrm{h}}\:=\:\frac{\mathrm{2}}{\pi}\:.\:\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\mathrm{2}.\:\mathrm{2}^{\mathrm{sin}\:\mathrm{h}} .\mathrm{cos}\:\mathrm{h}}{\mathrm{1}} \\ $$$$\:=\:\frac{\mathrm{2ln}\:\mathrm{2}}{\pi}\:=\:\pi^{−\mathrm{1}} .\mathrm{ln}\:\mathrm{4}\: \\ $$
Answered by Dwaipayan Shikari last updated on 08/Feb/21
lim_(x→(π/2)) ((2^(−cos(x)) −1)/(x(x−(π/2))))=((2^(−sin((π/2)−x)) −1)/(x(x−(π/2))))=((1−((π/2)−x)log(2)−1)/((π/2)(x−(π/2)))) =((2log(2))/π)=((log(4))/π) lim_(x→0) a^x =1+xlog(a)
$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\frac{\mathrm{2}^{−{cos}\left({x}\right)} −\mathrm{1}}{{x}\left({x}−\frac{\pi}{\mathrm{2}}\right)}=\frac{\mathrm{2}^{−{sin}\left(\frac{\pi}{\mathrm{2}}−{x}\right)} −\mathrm{1}}{{x}\left({x}−\frac{\pi}{\mathrm{2}}\right)}=\frac{\mathrm{1}−\left(\frac{\pi}{\mathrm{2}}−{x}\right){log}\left(\mathrm{2}\right)−\mathrm{1}}{\frac{\pi}{\mathrm{2}}\left({x}−\frac{\pi}{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{2}{log}\left(\mathrm{2}\right)}{\pi}=\frac{{log}\left(\mathrm{4}\right)}{\pi}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{a}^{{x}} =\mathrm{1}+{xlog}\left({a}\right) \\ $$
Answered by mathmax by abdo last updated on 08/Feb/21
f(x)=((2^(−cosx) −1)/(x(x−(π/2)))) we do the changement x−(π/2)=t(so t→o) and f(x)=f(t+(π/2)) =((2^(sint) −1)/(t(t+(π/2)))) we have 2^(sint) =e^(sint ln(2)) sint ∼t−(t^3 /6) ⇒e^(ln(2)sint) ∼e^(ln(2)(t−(t^3 /6))) ∼1+ln(2)(t−(t^3 /6)) ⇒ f(t+(π/2))∼((ln(2)(t−(t^3 /6)))/(t(t+(π/2)))) =((ln(2)(1−(t^2 /6)))/(t+(π/2)))→((2ln(2))/π) (t→0) ⇒ lim_(x→(π/2)) f(x)=((2ln(2))/π)
$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{2}^{−\mathrm{cosx}} −\mathrm{1}}{\mathrm{x}\left(\mathrm{x}−\frac{\pi}{\mathrm{2}}\right)}\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{x}−\frac{\pi}{\mathrm{2}}=\mathrm{t}\left(\mathrm{so}\:\mathrm{t}\rightarrow\mathrm{o}\right)\:\mathrm{and} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{t}+\frac{\pi}{\mathrm{2}}\right)\:=\frac{\mathrm{2}^{\mathrm{sint}} −\mathrm{1}}{\mathrm{t}\left(\mathrm{t}+\frac{\pi}{\mathrm{2}}\right)}\:\mathrm{we}\:\mathrm{have}\:\mathrm{2}^{\mathrm{sint}} \:=\mathrm{e}^{\mathrm{sint}\:\mathrm{ln}\left(\mathrm{2}\right)} \\ $$$$\mathrm{sint}\:\sim\mathrm{t}−\frac{\mathrm{t}^{\mathrm{3}} }{\mathrm{6}}\:\Rightarrow\mathrm{e}^{\mathrm{ln}\left(\mathrm{2}\right)\mathrm{sint}} \:\sim\mathrm{e}^{\mathrm{ln}\left(\mathrm{2}\right)\left(\mathrm{t}−\frac{\mathrm{t}^{\mathrm{3}} }{\mathrm{6}}\right)} \:\sim\mathrm{1}+\mathrm{ln}\left(\mathrm{2}\right)\left(\mathrm{t}−\frac{\mathrm{t}^{\mathrm{3}} }{\mathrm{6}}\right)\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{t}+\frac{\pi}{\mathrm{2}}\right)\sim\frac{\mathrm{ln}\left(\mathrm{2}\right)\left(\mathrm{t}−\frac{\mathrm{t}^{\mathrm{3}} }{\mathrm{6}}\right)}{\mathrm{t}\left(\mathrm{t}+\frac{\pi}{\mathrm{2}}\right)}\:=\frac{\mathrm{ln}\left(\mathrm{2}\right)\left(\mathrm{1}−\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{6}}\right)}{\mathrm{t}+\frac{\pi}{\mathrm{2}}}\rightarrow\frac{\mathrm{2ln}\left(\mathrm{2}\right)}{\pi}\:\left(\mathrm{t}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}} \:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{2ln}\left(\mathrm{2}\right)}{\pi} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *