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Question Number 203651 by Davidtim last updated on 24/Jan/24
lim_(x→1) (2−x)^(tan(((πx)/2))) =?
$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\mathrm{2}−{x}\right)^{{tan}\left(\frac{\pi{x}}{\mathrm{2}}\right)} =? \\ $$
Answered by Mathspace last updated on 24/Jan/24
changement 1−x=t give f(x)=(2−x)^(tan(((πx)/2))) =f(1−t)=(2−(1−t))^(tan((π/2)(1−t))) =(1+t)^(tan((π/2)−((πt)/2))) =(1+t)^(1/(tan(((πt)/2)))) (x→1⇔t→0) ln(f(1−t))=(1/(tan(((πt)/2))))ln(1+t) ∼(2/(πt))×t (t∈V(0) =(2/π) so lim(ln(f(1−t))=(2/π) ⇒ lim_(t→0) f(1−t)=e^(2/π) ⇒lim_(x→1) f(x) =e^(2/π)
$${changement}\:\mathrm{1}−{x}={t}\:{give} \\ $$$${f}\left({x}\right)=\left(\mathrm{2}−{x}\right)^{{tan}\left(\frac{\pi{x}}{\mathrm{2}}\right)} \\ $$$$={f}\left(\mathrm{1}−{t}\right)=\left(\mathrm{2}−\left(\mathrm{1}−{t}\right)\right)^{{tan}\left(\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−{t}\right)\right)} \\ $$$$=\left(\mathrm{1}+{t}\right)^{{tan}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi{t}}{\mathrm{2}}\right)} \\ $$$$=\left(\mathrm{1}+{t}\right)^{\frac{\mathrm{1}}{{tan}\left(\frac{\pi{t}}{\mathrm{2}}\right)}} \:\:\:\:\:\:\:\left({x}\rightarrow\mathrm{1}\Leftrightarrow{t}\rightarrow\mathrm{0}\right) \\ $$$${ln}\left({f}\left(\mathrm{1}−{t}\right)\right)=\frac{\mathrm{1}}{{tan}\left(\frac{\pi{t}}{\mathrm{2}}\right)}{ln}\left(\mathrm{1}+{t}\right) \\ $$$$\sim\frac{\mathrm{2}}{\pi{t}}×{t}\:\:\:\:\left({t}\in{V}\left(\mathrm{0}\right)\right. \\ $$$$=\frac{\mathrm{2}}{\pi}\:{so} \\ $$$${lim}\left({ln}\left({f}\left(\mathrm{1}−{t}\right)\right)=\frac{\mathrm{2}}{\pi}\:\Rightarrow\right. \\ $$$${lim}_{{t}\rightarrow\mathrm{0}} {f}\left(\mathrm{1}−{t}\right)={e}^{\frac{\mathrm{2}}{\pi}} \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{1}} \:{f}\left({x}\right)\:={e}^{\frac{\mathrm{2}}{\pi}} \\ $$
Answered by MM42 last updated on 24/Jan/24
1−x=u ⇒lim_(u→0) (1+u)^(cot((π/2)u)) ⇒lim_(u→0) ((1+u)^(1/u) )^((u/(sin ((πu)/2)))×cos((πu)/2)) =e^(2/π) ✓
$$\mathrm{1}−{x}={u} \\ $$$$\Rightarrow{lim}_{{u}\rightarrow\mathrm{0}} \:\left(\mathrm{1}+{u}\right)^{{cot}\left(\frac{\pi}{\mathrm{2}}{u}\right)} \\ $$$$\Rightarrow{lim}_{{u}\rightarrow\mathrm{0}} \:\left(\left(\mathrm{1}+{u}\right)^{\frac{\mathrm{1}}{{u}}} \right)^{\frac{{u}}{{sin}\:\frac{\pi{u}}{\mathrm{2}}}×{cos}\frac{\pi{u}}{\mathrm{2}}} \\ $$$$={e}^{\frac{\mathrm{2}}{\pi}} \:\:\checkmark \\ $$
Answered by Calculusboy last updated on 26/Jan/24
Solution: by sub directly,we get (1^∞ )indeterminant let y=(2−x)^(tan(((𝛑x)/2))) (apply log to both sides) logy=log(2−x)^(tan(((𝛑x)/2))) logy=lim_(x→1) tan(((𝛑x)/2))log(2−x) logy=lim_(x→1) (((d/dx)[log(2−x)])/((d/dx)[(1/(tan(((𝛑x)/2))))])) ⇒ logy=lim_(x→1) (((−1)/(2−x))/((d/dx)[cot(((𝛑x)/2))])) logy=lim_(x→1) (((−1)/(2−x))/([−(𝛑/2)cosec^2 (((𝛑x)/2))])) logy=(((−1)/1)/(−(𝛑/2)lim_(x→1) [cosec(((𝛑x)/2))]^2 )) logy=((−1)/(−(𝛑/2)∙1)) ⇒ logy=(2/𝛑) y=e^(2/𝛑) ∴lim_(x→1) (2−x)^(tan(((𝛑x)/2))) =e^(2/𝛑)
$$\boldsymbol{{Solution}}:\:\boldsymbol{{by}}\:\boldsymbol{{sub}}\:\boldsymbol{{directly}},\boldsymbol{{we}}\:\boldsymbol{{get}}\:\left(\mathrm{1}^{\infty} \right)\boldsymbol{{indeterminant}} \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{y}}=\left(\mathrm{2}−\boldsymbol{{x}}\right)^{\boldsymbol{{tan}}\left(\frac{\boldsymbol{\pi{x}}}{\mathrm{2}}\right)} \:\:\:\left(\boldsymbol{{apply}}\:\boldsymbol{{log}}\:\boldsymbol{{to}}\:\boldsymbol{{both}}\:\boldsymbol{{sides}}\right) \\ $$$$\boldsymbol{{logy}}=\boldsymbol{{log}}\left(\mathrm{2}−\boldsymbol{{x}}\right)^{\boldsymbol{{tan}}\left(\frac{\boldsymbol{\pi{x}}}{\mathrm{2}}\right)} \\ $$$$\boldsymbol{{logy}}=\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{1}} {\boldsymbol{{m}tan}}\left(\frac{\boldsymbol{\pi{x}}}{\mathrm{2}}\right)\boldsymbol{{log}}\left(\mathrm{2}−\boldsymbol{{x}}\right) \\ $$$$\boldsymbol{{logy}}=\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{1}} {\boldsymbol{{m}}}\frac{\frac{\boldsymbol{{d}}}{\boldsymbol{{dx}}}\left[\boldsymbol{{log}}\left(\mathrm{2}−\boldsymbol{{x}}\right)\right]}{\frac{\boldsymbol{{d}}}{\boldsymbol{{dx}}}\left[\frac{\mathrm{1}}{\boldsymbol{{tan}}\left(\frac{\boldsymbol{\pi{x}}}{\mathrm{2}}\right)}\right]}\:\:\:\Rightarrow\:\:\boldsymbol{{logy}}=\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{1}} {\boldsymbol{{m}}}\frac{\frac{−\mathrm{1}}{\mathrm{2}−\boldsymbol{{x}}}}{\frac{\boldsymbol{{d}}}{\boldsymbol{{dx}}}\left[\boldsymbol{{cot}}\left(\frac{\boldsymbol{\pi{x}}}{\mathrm{2}}\right)\right]} \\ $$$$\boldsymbol{{logy}}=\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{1}} {\boldsymbol{{m}}}\frac{\frac{−\mathrm{1}}{\mathrm{2}−\boldsymbol{{x}}}}{\left[−\frac{\boldsymbol{\pi}}{\mathrm{2}}\boldsymbol{{cosec}}^{\mathrm{2}} \left(\frac{\boldsymbol{\pi{x}}}{\mathrm{2}}\right)\right]} \\ $$$$\boldsymbol{{logy}}=\frac{\frac{−\mathrm{1}}{\mathrm{1}}}{−\frac{\boldsymbol{\pi}}{\mathrm{2}}\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{1}} {\boldsymbol{{m}}}\left[\boldsymbol{{cosec}}\left(\frac{\boldsymbol{\pi{x}}}{\mathrm{2}}\right)\right]^{\mathrm{2}} } \\ $$$$\boldsymbol{{logy}}=\frac{−\mathrm{1}}{−\frac{\boldsymbol{\pi}}{\mathrm{2}}\centerdot\mathrm{1}}\:\:\Rightarrow\:\:\boldsymbol{{logy}}=\frac{\mathrm{2}}{\boldsymbol{\pi}} \\ $$$$\boldsymbol{{y}}=\boldsymbol{{e}}^{\frac{\mathrm{2}}{\boldsymbol{\pi}}} \\ $$$$\therefore\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{1}} {\boldsymbol{{m}}}\left(\mathrm{2}−\boldsymbol{{x}}\right)^{\boldsymbol{{tan}}\left(\frac{\boldsymbol{\pi{x}}}{\mathrm{2}}\right)} =\boldsymbol{{e}}^{\frac{\mathrm{2}}{\boldsymbol{\pi}}} \\ $$$$ \\ $$

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