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Question Number 109191 by abdomsup last updated on 21/Aug/20
find lim_(x→(π/2)) (sinx)^(ln∣x−(π/2)∣)
$${find}\:\:{lim}_{{x}\rightarrow\frac{\pi}{\mathrm{2}}} \:\:\:\left({sinx}\right)^{{ln}\mid{x}−\frac{\pi}{\mathrm{2}}\mid} \\ $$
Commented by bemath last updated on 22/Aug/20
set x = (π/2)+ z L= lim_(z→0) (sin ((π/2)+z))^(ln ∣z∣) =lim_(z→0) (cos z)^(ln ∣z∣) ln L = lim_(z→0) ln ∣z∣ (cos z) ln L = lim_(z→0) ((cos z)/(1/(ln z))) = lim_(z→0) ((−sin z)/([((−(1/z))/(ln^2 (z)))])) ln L=lim_(z→0) sin z .((ln^2 z)/(1/z))=lim_(z→0) z.sin z.ln^2 z ln L = 0 ⇒ L = e^0 = 1
$${set}\:{x}\:=\:\frac{\pi}{\mathrm{2}}+\:{z} \\ $$$${L}=\:\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}+{z}\right)\right)^{\mathrm{ln}\:\mid{z}\mid} =\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{cos}\:{z}\right)^{\mathrm{ln}\:\mid{z}\mid} \\ $$$$\mathrm{ln}\:{L}\:=\:\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{ln}\:\mid{z}\mid\:\left(\mathrm{cos}\:{z}\right) \\ $$$$\mathrm{ln}\:{L}\:=\:\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cos}\:{z}}{\frac{\mathrm{1}}{\mathrm{ln}\:{z}}}\:=\:\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{sin}\:{z}}{\left[\frac{−\frac{\mathrm{1}}{{z}}}{\mathrm{ln}\:^{\mathrm{2}} \left({z}\right)}\right]} \\ $$$$\mathrm{ln}\:{L}=\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{sin}\:{z}\:.\frac{\mathrm{ln}\:^{\mathrm{2}} {z}}{\frac{\mathrm{1}}{{z}}}=\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{z}.\mathrm{sin}\:{z}.\mathrm{ln}\:^{\mathrm{2}} \:{z} \\ $$$$\mathrm{ln}\:{L}\:=\:\mathrm{0}\:\Rightarrow\:{L}\:=\:{e}^{\mathrm{0}} \:=\:\mathrm{1} \\ $$$$ \\ $$
Answered by mathmax by abdo last updated on 23/Aug/20
f(x) =(sinx)^(ln∣x−(π/2)∣) ⇒f(x) =e^(ln∣x−(π/2)∣ln(sinx)) changement x−(π/2)=t give f(x) =g(t) =e^(ln∣t∣ ln(sin((π/2)+t))) =e^(ln∣t∣ln(cost)) (x→(π/2) ⇒ t→0 ) we have ln(cost) =ln(cos∣t∣) ∼ln(1−((∣t∣^2 )/2)) ⇒ ∼−((∣t∣^2 )/2) ⇒ln∣t∣ln(cost) ∼−ln∣t∣ ((∣t∣^2 )/2) →0 (t→0) ⇒ lim_(t→0) g(t) =1 =lim_(x→(π/2)) f(x)
$$\mathrm{f}\left(\mathrm{x}\right)\:=\left(\mathrm{sinx}\right)^{\mathrm{ln}\mid\mathrm{x}−\frac{\pi}{\mathrm{2}}\mid} \:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{e}^{\mathrm{ln}\mid\mathrm{x}−\frac{\pi}{\mathrm{2}}\mid\mathrm{ln}\left(\mathrm{sinx}\right)} \:\mathrm{changement}\:\mathrm{x}−\frac{\pi}{\mathrm{2}}=\mathrm{t} \\ $$$$\mathrm{give}\:\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{g}\left(\mathrm{t}\right)\:=\mathrm{e}^{\mathrm{ln}\mid\mathrm{t}\mid\:\mathrm{ln}\left(\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}+\mathrm{t}\right)\right)} \:=\mathrm{e}^{\mathrm{ln}\mid\mathrm{t}\mid\mathrm{ln}\left(\mathrm{cost}\right)} \\ $$$$\left(\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}\:\Rightarrow\:\mathrm{t}\rightarrow\mathrm{0}\:\:\right)\:\mathrm{we}\:\mathrm{have}\:\mathrm{ln}\left(\mathrm{cost}\right)\:=\mathrm{ln}\left(\mathrm{cos}\mid\mathrm{t}\mid\right)\:\sim\mathrm{ln}\left(\mathrm{1}−\frac{\mid\mathrm{t}\mid^{\mathrm{2}} }{\mathrm{2}}\right)\:\Rightarrow \\ $$$$\sim−\frac{\mid\mathrm{t}\mid^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\mathrm{ln}\mid\mathrm{t}\mid\mathrm{ln}\left(\mathrm{cost}\right)\:\sim−\mathrm{ln}\mid\mathrm{t}\mid\:\frac{\mid\mathrm{t}\mid^{\mathrm{2}} }{\mathrm{2}}\:\rightarrow\mathrm{0}\:\left(\mathrm{t}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{t}\rightarrow\mathrm{0}} \mathrm{g}\left(\mathrm{t}\right)\:=\mathrm{1}\:=\mathrm{lim}_{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}} \:\mathrm{f}\left(\mathrm{x}\right) \\ $$$$ \\ $$

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